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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple Derivatives
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context.
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.Example A. Find the following derivatives.a. y = e cos2(x)
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.Example A. Find the following derivatives.a. y =e cos2(x) All derivatives are taken with respect to x. By the eu–Chain Rule with u = cos2(x),
8.
Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.Example A. Find the following derivatives.a. y =e cos2(x) All derivatives are taken with respect to x. By the eu–Chain Rule with u = cos2(x), y = e cos2(x) [cos2(x)]
9.
Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.Example A. Find the following derivatives.a. y =e cos2(x) All derivatives are taken with respect to x. By the eu–Chain Rule with u = cos2(x), y = e cos2(x) [cos2(x)] by the Power Chain Rule =e cos2 (x)2cos(x)[cos(x)]
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Summary of DerivativesIn this section we summarize the various types ofderivatives we have encountered and the algebraassociated with each type.Simple DerivativesBy simple derivative we mean there is only one inputand one output in the context. Hence if we only havey = y(x) or x = x(t), etc.. in such cases the variable thatthe derivative is taken with respect to is clear.Example A. Find the following derivatives.a. y =e cos2(x) All derivatives are taken with respect to x. By the eu–Chain Rule with u = cos2(x), y = e cos2(x) [cos2(x)] by the Power Chain Rule =e cos2 (x)2cos(x)[cos(x)] = –2ecos2 (x)cos(x)sin(x)
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Summary of Derivativesb. u = x2ey where y = cos(x)
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative i.e. take the derivative of y = x2ecos(x)
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.u = [x2]ey + x2[ey]
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.u = [x2]ey + x2[ey] = 2xey + x2ey[y]
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.u = [x2]ey + x2[ey] = 2xey + x2ey[y]Substitute y = cos(x) and y = –sin(x), we haveu = 2xecos(x) – x2sin(x)ecos(x)
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.u = [x2]ey + x2[ey] = 2xey + x2ey[y]Substitute y = cos(x) and y = –sin(x), we haveu = 2xecos(x) – x2sin(x)ecos(x) = xecos(x)(2 – xsin(x))
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Summary of Derivativesb. u = x2ey where y = cos(x)Assuming the derivative is taken with respect to x,there are two ways to track the derivation.We may plug in the cos(x) for y first then take thederivative or as we would proceed with theProduct Rule and track the derivation with y.u = [x2]ey + x2[ey] = 2xey + x2ey[y]Substitute y = cos(x) and y = –sin(x), we haveu = 2xecos(x) – x2sin(x)ecos(x) = xecos(x)(2 – xsin(x))The above derivation leads to implicit differentiation.
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v,
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on bothsides, v3 = 2 – uv][2u2 –
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Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on both sides,[2u2 – v3 = 2 – uv]4uu – 3v2 =
28.
Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on both sides,[2u2 – v3 = 2 – uv]4uu – 3v2 = –uv – uv
29.
Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on both sides,[2u2 – v3 = 2 – uv] 14uu – 3v2 = –uv – uv
30.
Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on both sides,[2u2 – v3 = 2 – uv] 14uu – 3v2 = –uv – uvu(4u + v) = 3v2 – u
31.
Summary of DerivativesImplicit DerivativesIn implicit differentiation, a two (or more)–variableequation is given, we assume one of them as theindependent variable, say v, and we are to find thederivative of the other variable with respect to v,hence the Chain Rules are needed for derivatives ofthe variable in the equations.Example B. Given that 2u2 – v3 = 2 – uv,a. Find the derivative of u with respect to v.Taking the derivative with respect to v on both sides,[2u2 – v3 = 2 – uv] 14uu – 3v2 = –uv – uvu(4u + v) = 3v2 – u 3v2 – uso u = 4u + v
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Summary of Derivativesb. Find the derivative of v with respect to u.
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uv
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4u
36.
Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4uv(u – 3v2) = – v – 4u so
37.
Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4uv(u – 3v2) = – v – 4u so – 4u – v 4u + vso v = u – 3v2 = 3v2 – u
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4uv(u – 3v2) = – v – 4u so – 4u – v 4u + vso v = u – 3v2 = 3v2 – uIn the differential–notation, the reciprocal relation ofthese derivatives becomes clear.
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4uv(u – 3v2) = – v – 4u so – 4u – v 4u + vso v = u – 3v2 = 3v2 – uIn the differential–notation, the reciprocal relation ofthese derivatives becomes clear. du 3v2 – u dv = 4u + v dv = 4u + v du 3v2 – u
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Summary of Derivativesb. Find the derivative of v with respect to u.Taking the derivative with respect to u on bothsides, v3 = 2 – uv][2u2 –4u – 3v2v = –v – uvuv – 3v2v = –v – 4uv(u – 3v2) = – v – 4u so – 4u – v 4u + vso v = u – 3v2 = 3v2 – uIn the differential–notation, the reciprocal relation ofthese derivatives becomes clear. du 3v2 – u dv = 4u + v dv = 4u + v du 3v2 – uThis follows the fact that the slopes at two diagonallyreflected points on the graphs of a pair of inversefunctions are the reciprocal of each other.
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Related RatesRelated Rates DerivativesIn related rates problems all variables are assumed tobe functions of a single variable t, usually for time,
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Related RatesRelated Rates DerivativesIn related rates problems all variables are assumed tobe functions of a single variable t, usually for time,and all the derivatives (#) are taken with respect to t.
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Related RatesRelated Rates DerivativesIn related rates problems all variables are assumed tobe functions of a single variable t, usually for time,and all the derivatives (#) are taken with respect to t.Hence the derivatives of all of the variables in thecontext require the Chain Rules.
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Related RatesRelated Rates DerivativesIn related rates problems all variables are assumed tobe functions of a single variable t, usually for time,and all the derivatives (#) are taken with respect to t.Hence the derivatives of all of the variables in thecontext require the Chain Rules. In a related rateproblem, the defining relation usually is formula suchas the Distance Formula, or any formula from anyscientific discipline that relates multiple variables whereeach variable is a function in t.
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Related RatesRelated Rates DerivativesIn related rates problems all variables are assumed tobe functions of a single variable t, usually for time,and all the derivatives (#) are taken with respect to t.Hence the derivatives of all of the variables in thecontext require the Chain Rules. In a related rateproblem, the defining relation usually is formula suchas the Distance Formula, or any formula from anyscientific discipline that relates multiple variables whereeach variable is a function in t.Often in these problems, the explicit formulas in terms tare bypassed. Instead the relations and values of thevariables and their rates with respect to t are providedwith numeric inputs.
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0 or v = dv/dt = 6
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0 or v = dv/dt = 6The Geometry of Derivatives
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0 or v = dv/dt = 6The Geometry of DerivativesAt a generic point x, the derivative f(x) of thefunction f(x) gives the “slope” at the point (x, f(x))on the graph of y = f(x).
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0 or v = dv/dt = 6The Geometry of DerivativesAt a generic point x, the derivative f(x) of thefunction f(x) gives the “slope” at the point (x, f(x))on the graph of y = f(x). The existence of this slopemeans the graph is smooth at that point.
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Summary of DerivativesExample C. Given that 2u2 – v3 + 2uv = 2 where uand v are functions in t, given that du/dt = –3 at thepoint (u = 1, v = 0), find dv/dt at that instant.Taking the derivative with respect to t using theprime–notation,[2u2 – v3 + 2uv = 2]4uu – 3v2v + 2uv + 2uv = 0Substitute u = –3 at u = 1, v = 0, we get–12 + 2v = 0 or v = dv/dt = 6The Geometry of DerivativesAt a generic point x, the derivative f(x) of thefunction f(x) gives the “slope” at the point (x, f(x))on the graph of y = f(x). The existence of this slopemeans the graph is smooth at that point.
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y= f(x)point P is f(u) and the f(u).slope at the point Q is f(v) Q P (v, f(v))as shown. (u, f(u)) x
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y=point P is f(u) and the f(u). f(x)slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y=point P is f(u) and the f(u). f(x)slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y= f(x)point P is f(u) and the f(u).slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x).
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y=point P is f(u) and the f(u). f(x)slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x). These different cases may be identified bythe 2nd derivative f (x) as shown.
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y=point P is f(u) and the f(u). f(x)slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x). These different cases may be identified bythe 2nd derivative f (x) as shown. (x, f(x)) f (x) < 0 a max.
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Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y=point P is f(u) and the f(u). f(x)slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x). These different cases may be identified bythe 2nd derivative f (x) as shown. (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 a max. a min.
63.
Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y= f(x)point P is f(u) and the f(u).slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x). These different cases may be identified bythe 2nd derivative f (x) as shown. (x, f(x)) (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 f (x)<0 f (x)>0 a max. a min. going up flat pt.
64.
Derivatives and GraphsHence the slope at the The slope is f(v) . The slope is y= f(x)point P is f(u) and the f(u).slope at the point Q is f(v) Q Pas shown. (u, f(u)) (v, f(v))The signs of f(x) indicate the xups and downs of the curve.I. If f (x) = 0 the tangent line is flat at (x, f(x)).There are four possible shapes of the graph ofy = f(x). These different cases may be identified bythe 2nd derivative f (x) as shown. (x, f(x)) (x, f(x)) (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 f (x)<0 f (x)>0 f (x)>0 f (x)<0 a max. a min. going up flat pt. going down flat pt.
65.
Derivatives and GraphsII. If f (x) > 0, the slope is positive at (x, f(x)) so thegraph is going uphill. There are four possibilities forthe graph. The four cases and their 2nd derivativef (x) are shown below. (x, f(x)) (x, f(x)) (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 f (x)<0 f (x)>0 f (x)>0 f (x)<0
66.
Derivatives and GraphsII. If f (x) > 0, the slope is positive at (x, f(x)) so thegraph is going uphill. There are four possibilities forthe graph. The four cases and their 2nd derivativef (x) are shown below. (x, f(x)) (x, f(x)) (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 f (x)<0 f (x)>0 f (x)>0 f (x)<0III. If f (x) < 0, i.e. the graph is downhill at (x, f(x)).The four possible shapes are shown here. (x, f(x)) (x, f(x)) (x, f(x)) (x, f(x)) f (x) < 0 f (x) > 0 f (x)<0 f (x)>0 f (x)>0 f (x)<0
67.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).
68.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3]
69.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v
70.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3v
71.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.
72.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we get
73.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we getSo v = (2 + 3v2)2
74.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we getSo v = (2 + 3v2)(4) – 4u(6vv) (2 + 3v2)2
75.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we getSo v = (2 + 3v2)(4) – 4u(6vv) (2 + 3v2)2At u = 0, v = –1, and v = 0 we getv > 0.
76.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we getSo v = (2 + 3v2)(4) – 4u(6vv) (2 + 3v2)2At u = 0, v = –1, and v = 0 we getv > 0. From these, we conclude that(0, –1) must be a minimum.
77.
Summary of DerivativesExample D. Given that 2u2 – v3 = 2v + 3 draw thegraph of v = v(u) near the point (u, v) = (0, –1).Take the derivative with respect to u to find dv/du,[2u2 – v3 = 2v + 3] 4u – 3v2v = 2v So v = 2 4u 2 + 3vAt u = 0, v = –1, we get v = 0.Take the derivative of v with respect to u again, we getSo v = (2 + 3v2)(4) – 4u(6vv) v (2 + 3v2)2At u = 0, v = –1, and v = 0 we get uv > 0. From these, we conclude that (0,–1)(0, –1) must be a minimum. v(0) = 0 v(0) > 0
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