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DERIVATIVES implicit function.pptx
- 1. DERIVATIVES Derivatives of implicit functions
- 2. Find derivative of π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = 0 β’ π ππ₯ π₯2 + π¦2 + 2ππ₯ + 2ππ¦ + π = π0 ππ₯ β’ ππ₯2 ππ₯ + ππ¦2 ππ₯ + π2ππ₯ ππ₯ + π2ππ¦ ππ₯ + ππ ππ₯ =0 β’ 2π₯ + 2π¦ ππ¦ ππ₯ + 2π 1 + 2π ππ¦ ππ₯ = 0 β’ 2π¦ ππ¦ ππ₯ + 2π ππ¦ ππ₯ = β2π₯ β 2π β’ ππ¦ ππ₯ 2π¦ + 2π = β2π₯ β 2π β’ ππ¦ ππ₯ = β2π₯β2π 2π¦+2π = 2(βπ₯βπ) 2(π¦+π) = βπ₯βπ π¦+π
- 3. Find derivative of ππ₯2 + 2βπ₯π¦ + ππ¦2 + 2ππ₯ + 2ππ¦ + π = 0 β’ π ππ₯ ππ₯2 + 2βπ₯π¦ + ππ¦2 + 2ππ₯ + 2ππ¦ + π = π0 ππ₯ β’ πππ₯2 ππ₯ + π2βπ₯π¦ ππ₯ + πππ¦2 ππ₯ + π2ππ₯ ππ₯ + π2ππ¦ ππ₯ + ππ ππ₯ =0 β’ 2ππ₯ + 2β ππ₯π¦ ππ₯ + 2ππ¦ ππ¦ ππ₯ + 2π 1 + 2π ππ¦ ππ₯ = 0 β’ 2ππ₯ + 2β π¦ ππ₯ ππ₯ + π₯ ππ¦ ππ₯ + 2by ππ¦ ππ₯ + 2π + 2π ππ¦ ππ₯ = 0 β’ 2ππ₯ + 2β π¦ 1 + π₯ ππ¦ ππ₯ + 2ππ¦ ππ¦ ππ₯ + 2g + 2f ππ¦ ππ₯ = 0 β’ 2ππ₯ + 2βπ¦ + 2βπ₯ ππ¦ ππ₯ + 2ππ¦ ππ¦ ππ₯ + 2π + 2π ππ¦ ππ₯ = 0 β’ 2βπ₯ ππ¦ ππ₯ + 2ππ¦ ππ¦ ππ₯ + 2π ππ¦ ππ₯ = β2ππ₯ β 2βπ¦ β 2π β’ ππ¦ ππ₯ 2βπ₯ + 2ππ¦ + 2π = β2ππ₯ β 2βπ¦ β 2π β’ ππ¦ ππ₯ = β2ππ₯β2βπ¦β2π 2βπ₯+2ππ¦+2π β’ ππ¦ ππ₯ = β2(ππ₯ββπ¦βπ) 2(βπ₯+ππ¦+π) = β(ππ₯ββπ¦βπ) (βπ₯+ππ¦+π)
- 4. Find derivative of π₯2 π2 + π¦2 π2 = 1 β’ π ππ ππ π π + π ππ ππ π π = π π π π β’ ππ ππ+ ππ ππ π π π π = π β’ ππ ππ π π π π =- ππ ππ β’ π π π π = - ππ ππ Γ ππ ππ β’ π π π π = βπππ πππ
- 5. Find derivative of logarithmic functions β’ Sometimes a function is given as a product and quotient of number of functions in such condition we need to logarithm of both the side of function to differentiating it. β’ For example β’ π π₯ = π₯2 6π₯+7 9 3π₯+2 7 β’ π¦ = π₯π₯ β’ π¦ = π₯ππ₯ β’ π₯π¦ = π¦π₯ β’ π¦ = π₯ 3 2 1β4π₯ 7 5βπ₯ 7 2 15β8π₯ 1 2
- 6. Find derivative of (i) π¦ = π₯π₯ (ii)y=π₯ππ₯ β’ π¦ = π₯π₯ β’ Taking logarithm of both the side β’ ππππ¦ = ππππ₯π₯ β’ ππππ¦ = π₯ππππ₯ β’ πππππ¦ ππ₯ = ππ₯ππππ₯ ππ₯ β’ 1 π¦ Γ ππ¦ ππ₯ = π₯ πππππ₯ ππ₯ + ππππ₯ ππ₯ ππ₯ β’ 1 π¦ Γ ππ¦ ππ₯ = π₯ 1 π₯ + ππππ₯ 1 β’ 1 π¦ Γ ππ¦ ππ₯ = (1 + ππππ₯) β’ ππ¦ ππ₯ = π¦(1 + ππππ₯)=π₯π₯ 1 + ππππ₯ β’ y=π₯ππ₯ β’ Taking logarithm of both the side β’ ππππ¦ = πππ π₯ππ₯ (πππππ = πππππ) β’ β΄ ππππ¦ = ππ₯ ππππ₯ β’ πππππ¦ ππ₯ = πππ₯ππππ₯ ππ₯ β’ 1 π¦ Γ ππ¦ ππ₯ = ππππ₯ πππ₯ ππ₯ + ππ₯ πππππ₯ ππ₯ β’ 1 π¦ Γ ππ¦ ππ₯ = ππππ₯ ππ₯ + ππ₯ 1 π₯ β’ 1 π¦ Γ ππ¦ ππ₯ = ππ₯ (ππππ₯ + 1 π₯ ) β’ ππ¦ ππ₯ = π¦ ππ₯(ππππ₯ + 1 π₯ )=π₯ππ₯ ππ₯(ππππ₯ + 1 π₯ )
- 7. Find derivative of π¦ = π₯+8 3βπ₯ 2 5 β’ π¦ = π+π πβπ π π β’ Taking logarithm of both the side β’ ππππ = πππ π+π πβπ π π β’ ππππ = π π πππ π+π πβπ πππππ = πππππ β’ π ππππ π π = π π π₯π¨π π + π β π₯π¨π (π β π) (πππ π π =ππππ β ππππ) β’ π π Γ π π π π = π π π π π πππ π + π β πππ(π β π) β’ π π Γ π π π π = π π π π π πππ π + π β π π π πππ π β π β’ π π Γ π π π π = π π π (π+π) π π π (π + π) β π (πβπ) π π π (π β π) πππππ ππππ, π ππππ π π = π π β’ π π Γ π π π π = π π π (π+π) (π) β π (πβπ) (βπ) β’ π π Γ π π π π = π π π π+π + π πβπ β’ 1 π¦ Γ ππ¦ ππ₯ = 2 5 3βπ₯ +(π₯+8) (π₯+8)(3βπ₯) β’ = 2 5 3βπ₯+π₯+8 (π₯+8)(3βπ₯) β’ = 2 5 11 (π₯+8)(3βπ₯) β’ 1 π¦ Γ ππ¦ ππ₯ = 22 5(π₯+8)(3βπ₯) β’ ππ¦ ππ₯ =π¦ 22 5(π₯+8)(3βπ₯) β’ ππ¦ ππ₯ = π₯+8 3βπ₯ 2 5 22 5(π₯+8)(3βπ₯)
- 8. Find derivative of π₯π¦ = π¦π₯ β’ π₯π¦ = π¦π₯ β’ Taking logarithm of both the sides β’ ππππ₯π¦ = ππππ¦π₯ β’ π¦ππππ₯ = π₯ππππ¦ β’ π ππ₯ π¦ππππ₯ = π ππ₯ π₯ππππ¦ (applying law of multiplication ) β’ ππππ₯ Γ ππ¦ ππ₯ + π¦ π ππ₯ ππππ₯=ππππ¦ π ππ₯ π₯ + π₯ π ππ₯ ππππ¦ β’ ππππ₯ Γ ππ¦ ππ₯ + π¦ 1 π₯ =ππππ¦ 1 + π₯ 1 π¦ ππ¦ ππ₯ β’ ππππ₯ ππ¦ ππ₯ + π¦ π₯ = ππππ¦ + π₯ π¦ ππ¦ ππ₯ β’ ππππ₯ ππ¦ ππ₯ β π₯ π¦ ππ¦ ππ₯ =logy β π¦ π₯ β’ ππ¦ ππ₯ (ππππ₯ β π₯ π¦ ) =logy β π¦ π₯ β’ ππ¦ ππ₯ π¦ππππ₯βπ₯ π¦ = π₯ππππ¦βπ¦ π₯ β’ ππ¦ ππ₯ = π¦(π₯ππππ¦βπ¦) π₯(π¦ππππ₯βπ₯)
- 9. Find derivative of π¦ = π₯ 3 2 1β4π₯ 7 5βπ₯ 7 2 15β8π₯ 1 2 β’ π¦ = π₯ 3 2 1β4π₯ 7 5βπ₯ 7 2 15β8π₯ 1 2 β’ Taking log of both the side β’ ππππ¦ = πππ π₯ 3 2 1β4π₯ 7 5βπ₯ 7 2 15β8π₯ 1 2 β’ ππππ¦ = πππ π₯ 3 2 +πππ 1 β 4π₯ 7 β πππ 5 β π₯ 7 2 + πππ 15 β 8π₯ 1 2 β’ ππππ¦ = 3 2 ππππ₯ + 7log(1 β 4π₯) β 7 2 log 5 β π₯ + 1 2 log(15 β 8π₯) β’ π ππ₯ ππππ¦ = π ππ₯ 3 2 ππππ₯ + 7log(1 β 4π₯) β 7 2 log 5 β π₯ + 1 2 log(15 β 8π₯) β’ 1 π¦ ππ¦ ππ₯ = 3 2 Γ 1 π₯ + 7 1 1β4π₯ π(1β4π₯) ππ₯ β 7 2 1 (5βπ₯) π(5βπ₯) ππ₯ + 1 2 1 (15β8π₯) π(15β8π₯) ππ₯ β’ 1 π¦ ππ¦ ππ₯ = 3 2π₯ + 7(β4) 1β4π₯ β 7 2 5βπ₯ (β1) + 1 2 15β8π₯ (β8) β’ ππ¦ ππ₯ = π₯ 3 2 1β4π₯ 7 5βπ₯ 7 2 15β8π₯ 1 2 3 2π₯ + (β28) 1β4π₯ β β7 2 5βπ₯ + β8 2 15β8π₯
- 10. Derivative of parametric equations β’ If y =f(t) and x=g(t) then ππ¦ ππ₯ = ππ¦ ππ‘ ππ₯ ππ‘ here t is parameter and equations are known as parametric equations β’ For e.g. x=ππ‘2 and y = 2ππ‘ β’ Y=π₯π₯π₯π₯π₯βββββ β’ Y=ππππ‘ + 1 πππ π₯ = π‘ππ‘ β’ Differentiate π₯2ππ₯ π€ππ‘β ππππππ‘ π‘π log π₯ β’ Differentiate π₯ππππ₯ π€ππ‘β πππ ππππ‘ π‘π 3π₯π₯3 β’ Y= π₯ + π₯ + π₯ + π₯ + β― β¦ β¦ β¦ β¦ β¦ β¦ β¦ . . β
- 11. Application of derivatives : β’ Study of derivative is important in various fields like β’ Economics β’ Business β’ Psychology β’ Many problems of trade and commerce β’ Maximization of profit and minimization of cost and loss . β’ To obtain marginal cost and marginal revenue β’ To obtain elasticity of demand and supply β’ To obtain maximum and minimum value of any function and stationary values
- 12. Homogeneous functions of two variables β’ A function is called homogeneous if the degree of all the terms are equal then only it can be represent in the form of π = πππ π π , or π = πππ π π β’ A function u(x , y) is called homogeneous of degree n if u(x , y) is expressed in the form of β’ π = πππ π π , or π = πππ π π β’ For example : π+π π+ π because it can be represented as π = π π π π+ π π π+ π π = π π ππ π π β’π = π+π π+ π β’ = π (π+π) π π ( π+ π) π β’ = π₯ π₯ π₯ + π¦ π₯ π₯ π₯ π₯ + π¦ π₯ = π₯ 1+ π¦ π₯ π₯ 1 2 1+ π¦ π₯ β’ =π₯1β 1 2 1+ π¦ π₯ 1+ π¦ π₯ = π₯ 1 2 1+ π¦ π₯ 1+ π¦ π₯ =π₯ 1 2 π π¦ π₯
- 13. Eulerβs theorem β’ Homogeneity of degree one is often called linear homogeneity . β’ linear homogeneous function U=x+2y β’ An important property of homogeneous function is given by Euler's theorem. β’ It is developed by swiss mathematician Leonhard Euler β’ It is mathematical relationship that applies to homogeneous function. β’ If production function is homogeneous and one degree one and factors of production are paid equal to marginal products (income =payment to factor (rent , wages, profit ,interest )total product is exhausted with no surplus(profit) and deficit(loss). β’ With the help of this theorem we can determine ,labor , capital for maximum production
- 14. Euler's theorem β’ If function u(x , y) is homogeneous function of degree n then β’ π₯ ππ’ ππ₯ + π¦ ππ’ ππ¦ = ππ’ πππ π₯2 π2π¦ ππ₯2 + 2π₯π¦ ππ’ ππ₯ππ¦ + π¦2 π2π₯ ππ¦2 = π π β 1 π’