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DIFFERENTIATION OF
EXPONENTIAL FUNCTIONS
OBJECTIVES:
• apply the properties of exponential
functions to simplify differentiation;
• differentiate functions involving exponential
functions; and
• solve problems involving differentiation of
exponential functions.
The EXPONENTIAL FUNCTION
.ylogx
aswrittenbealsomayayfunction,clogarithmi
ofinversetheisfunctionlexponentiatheSince
number.realaisxwhereaybydefinedis1,a
and0aa,basewithfunctionlexponentiaThe
a
x
x
=
=
=≠
>
.
.
nmanama.1 +=⋅








<
=
>−
=
nmif,m-na
1
nmif,1
nmif,nma
na
ma
.2
( ) mna
nma.3 =
( ) nbnanab.4 =
nb
nan
b
a
.5 =





0aprovided,10a.6 ≠=
n1ma
mn1
anma.7 











==
Laws of Exponents
xa.8 xloga
=
yxthenaaif.9 yx
==
DIFFERENTIATION FORMULA
Derivative of Exponential Function
The derivative of the exponential function for
any given base and any differentiable function of u.
( )
f(x)uwhere;
dx
du
e)e(
dx
d
f(x)uwhere;
dx
du
alna)a(
dx
d
uu
uu
==
=
==
:ebaseFor
:abasegivenanyFor
A. Find the derivative of each of the following
natural logarithm and simplify the result:
( )
2x3exf.1 =
( ) x21
exg.2 −
=
( ) x/12
ex4xh.3 =
( ) ( )
( )
2
2
x3
x3
xe6x'f
x6ex'f
=
=
( )
x212
2
ex'g x21
−
−
⋅= −
( ) ( ) ( )





+




 −
= x2e
x
1
ex4x'h x/1
2
x/12
EXAMPLE:
( )
x21
x21
x21
e
x'g
x21
−
−
•
−
−=
−
( ) ( )x21e4x'h x/1
+−=
( ) ( )1x2e4x'h x/1
−=
( )
x21
x21e
x'g
x21
−
−
−=
−
2
y
x2xxye.4 +=+
[ ] [ ] [ ] 0
2y
'yx1y
x21y'xyxye +
−
=+⋅+
2y
'xyy
x2yxye'yxyxe
−
=++
'xyy2xy2xye3y'yxye2xy −=++
2xy2xye3yyxxye2xy'y −−=+ 

















+
−−
=
xye2y1x
xye2yxy21y
'y
5x42x37y.5 +−=
( ) 




 +−+−= 5x42x3
dx
d
7ln5x42x37'y
( )[ ]4x67ln5x42x37'y −+−=
( )( ) 5x42x377ln2x32'y +−−=
( )
2x34lnxh.6 =
2x34
2x34
dx
d
)x('h






=
( )
( )
2x34
2x3
dx
d
4ln
2x34
x'h






=
( ) ( )[ ]x64lnx'h =
( ) 4lnx6x'h =
( )
2x34lnxh =
( ) 4ln2x3xh =
( ) ( ) 




= 2x
dx
d
4ln3x'h
( ) ( )[ ]x24ln3x'h =
( ) 4lnx6x'h =
OR
( ) ( ) 




 ++= 3x2e1xelogxG.6
( ) ( ) ( )3elog1elogxG x2x
+++=
( ) elog
3e
2e
elog
1e
e
x'G x2
x2
x
x
+
⋅
+
+
=
( ) ( ) ( )
( )( ) elog
3e1e
1ee23ee
x'G x2x
xx2x2x
++
+++
=
( )
( )( ) eloge
3e1e
e2e23e
x'G x
x2x
xx2x2
++
+++
=
( )
( )( ) eloge
3e1e
3e2e3
x'G x
x2x
xx2
++
++
=
( )
24
xx3
52xf.7 ⋅=
( ) ( ) ( )4224
x3xxx3
2
dx
d
55
dx
d
2x'f +=
( ) ( )[ ] ( )[ ]3x3xxx3
x122ln25x25ln52x'f
4224
+=
( ) [ ]2lnx65ln52x2x'f 2xx3 24
+=
( ) ( )x2lnx65ln52x'f 2x1x3 24
+= +
( )
24
xx3
52xf ⋅=
( ) ( )24
xx3
52lnxfln =
( )
24
xx3
5ln2lnxfln +=
( ) 5lnx2lnx3xfln 24
+=
( )
( )
( )[ ] ( )[ ]x25lnx42ln3
xf
x'f 3
+=
( )
( )
[ ]5ln2lnx6x2
xf
x'f 2
+=
[ ]5ln2lnx6x252)x('f 2xx3 24
+⋅⋅=
( ) ( )x5ln2lnx652x'f 2x1x3 24
+= +
OR
yx53.8 4yx
+=+
( ) ( ) 'yx4'y5ln53ln3 3yx
+=+
( )[ ] ( )3ln3x415ln5'y x3y
−=−
( )
( )[ ]15ln5
3ln3x4
'y y
x3
−
−
=
A. Find the derivative and simplify the result.
( ) 1x3x2
3xg.1 +−
=
( )
22
xlnx
exf.2 +
=
2e
e
y.3 x3
x4
+
=
( )
3
x22
2 3xlogxh.4 ⋅=
( )
2
x
5xG.1 =
2lnyxxeye.2 22yx
++=+
( ) ( )
2
X
1xxH.3 +=
1x2
e
y.4
1x2
+
=
+
( ) ( )x2x2
eelnxf.5 −
+=
B. Apply the appropriate formulas to obtain the
derivative of the given function and simplify.
EXERCISES:
Logarithmic Differentiation
Oftentimes, the derivatives of algebraic functions
which appear complicated in form (involving
products, quotients and powers) can be found
quickly by taking the natural logarithms of both
sides and applying the properties of logarithms
before differentiation. This method is called
logarithmic differentiation.
1. Take the natural logarithm of both sides and
apply the properties of logarithms.
2. Differentiate both sides and reduce the right
side to a single fraction.
3. Solve for y’ by multiplying the right side by y.
4. Substitute and simplify the result.
Steps in applying logarithmic differentiation.
Logarithmic differentiation is also applicable
whenever
the base and its power are both functions.
x
xyif
dx
dy
Find.1 =
xlnxyln
xlnyln x
=
=
Logarithmic differentiation is also applicable
whenever the base and its power are both functions.
(Variable to variable power.)
Example:
( ) ( )1xln1
x
1
x'y
y
1
+=
( ) x
xybutyxln1'y =→+=
( )( )x
xxln1'y +=∴
( )
( ) ( )1x2ln1xyln
1x2lnyln
1x
+−=
+=
−
( ) 1x
1x2yif
dx
dy
Find.2
−
+=
( ) ( ) ( )( )11x2ln2
1x2
1
1x'y
y
1
++
+
−=
( ) ( )1x2ln
1x2
1x2
'y
y
1
++
+
−
=
( ) ( ) ( ) 1-x
12xybuty1x2ln
1x2
1x2
'y +=→





++
+
−
=
( ) ( ) ( ) ( ) 1-x
12x
1x2
1x2ln1x21x2
'y +





+
+++−
=
( ) ( ) ( )[ ]( ) 1-1-x
12x1x2ln1x21x2'y ++++−=
( ) ( ) ( )[ ]( ) 2-x
12x1x2ln1x21x2'y ++++−=∴
( ) x
5x6y.3 +=
( )
5x6lnxyln
5x6lnyln
x
+=
+=






++





+






+
=
x2
1
5x6ln
5x62
6
5x6
1
xy'
y
1
( ) ( ) 1
2
x
5x6
x2
5x6ln5x6x6
y'
−
+




 +++
=∴
x2
5x6ln
5x6
x3
y'
y
1 +
+
+
=
( )
( )5x6x2
5x6ln5x6x6
y'
y
1
+
+++
=
( )
( )
( ) ( ) x
5x6ybuty
5x6x2
5x6ln5x6x6
y' +=⇒





+
+++
=
( )
( )
( ) x
5x6
5x6x2
5x6ln5x6x6
y' +





+
+++
=
( ) 1x
x34y.4
−
−=
( )
( )x34ln1xyln
x34lnyln
1x
−−=
−=
−
( ) ( ) 





−
−+





−
−
−=
1x2
1
x34ln3
x34
1
1x'y
y
1
( )
1x2
x34ln
x34
1x3
'y
y
1
−
−
+
−
−−
=
( ) ( ) ( )
( ) 1xx342
x34lnx341x6
'y
y
1
−−
−−+−−
=
( ) ( ) ( )
( )
( ) ( ) 1x
x34ybuty
1xx342
x34lnx341x6
y'
−
−=⇒





−−
−−+−−
=
( ) ( ) ( )
( )
( ) 1x
x34
1xx342
x34lnx341x6
'y
−
−





−−
−−+−−
=
( ) ( ) ( ) ( ) 11x
x34
1x2
x34lnx341x6
y'
−−
−





−
−−+−−
=∴

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Lesson 10 derivative of exponential functions

  • 2. OBJECTIVES: • apply the properties of exponential functions to simplify differentiation; • differentiate functions involving exponential functions; and • solve problems involving differentiation of exponential functions.
  • 4. . nmanama.1 +=⋅         < = >− = nmif,m-na 1 nmif,1 nmif,nma na ma .2 ( ) mna nma.3 = ( ) nbnanab.4 = nb nan b a .5 =      0aprovided,10a.6 ≠= n1ma mn1 anma.7             == Laws of Exponents xa.8 xloga = yxthenaaif.9 yx ==
  • 5. DIFFERENTIATION FORMULA Derivative of Exponential Function The derivative of the exponential function for any given base and any differentiable function of u. ( ) f(x)uwhere; dx du e)e( dx d f(x)uwhere; dx du alna)a( dx d uu uu == = == :ebaseFor :abasegivenanyFor
  • 6. A. Find the derivative of each of the following natural logarithm and simplify the result: ( ) 2x3exf.1 = ( ) x21 exg.2 − = ( ) x/12 ex4xh.3 = ( ) ( ) ( ) 2 2 x3 x3 xe6x'f x6ex'f = = ( ) x212 2 ex'g x21 − − ⋅= − ( ) ( ) ( )      +      − = x2e x 1 ex4x'h x/1 2 x/12 EXAMPLE: ( ) x21 x21 x21 e x'g x21 − − • − −= − ( ) ( )x21e4x'h x/1 +−= ( ) ( )1x2e4x'h x/1 −= ( ) x21 x21e x'g x21 − − −= −
  • 7. 2 y x2xxye.4 +=+ [ ] [ ] [ ] 0 2y 'yx1y x21y'xyxye + − =+⋅+ 2y 'xyy x2yxye'yxyxe − =++ 'xyy2xy2xye3y'yxye2xy −=++ 2xy2xye3yyxxye2xy'y −−=+                   + −− = xye2y1x xye2yxy21y 'y
  • 8. 5x42x37y.5 +−= ( )       +−+−= 5x42x3 dx d 7ln5x42x37'y ( )[ ]4x67ln5x42x37'y −+−= ( )( ) 5x42x377ln2x32'y +−−= ( ) 2x34lnxh.6 = 2x34 2x34 dx d )x('h       = ( ) ( ) 2x34 2x3 dx d 4ln 2x34 x'h       = ( ) ( )[ ]x64lnx'h = ( ) 4lnx6x'h = ( ) 2x34lnxh = ( ) 4ln2x3xh = ( ) ( )      = 2x dx d 4ln3x'h ( ) ( )[ ]x24ln3x'h = ( ) 4lnx6x'h = OR
  • 9. ( ) ( )       ++= 3x2e1xelogxG.6 ( ) ( ) ( )3elog1elogxG x2x +++= ( ) elog 3e 2e elog 1e e x'G x2 x2 x x + ⋅ + + = ( ) ( ) ( ) ( )( ) elog 3e1e 1ee23ee x'G x2x xx2x2x ++ +++ = ( ) ( )( ) eloge 3e1e e2e23e x'G x x2x xx2x2 ++ +++ = ( ) ( )( ) eloge 3e1e 3e2e3 x'G x x2x xx2 ++ ++ =
  • 10. ( ) 24 xx3 52xf.7 ⋅= ( ) ( ) ( )4224 x3xxx3 2 dx d 55 dx d 2x'f += ( ) ( )[ ] ( )[ ]3x3xxx3 x122ln25x25ln52x'f 4224 += ( ) [ ]2lnx65ln52x2x'f 2xx3 24 += ( ) ( )x2lnx65ln52x'f 2x1x3 24 += + ( ) 24 xx3 52xf ⋅= ( ) ( )24 xx3 52lnxfln = ( ) 24 xx3 5ln2lnxfln += ( ) 5lnx2lnx3xfln 24 += ( ) ( ) ( )[ ] ( )[ ]x25lnx42ln3 xf x'f 3 += ( ) ( ) [ ]5ln2lnx6x2 xf x'f 2 += [ ]5ln2lnx6x252)x('f 2xx3 24 +⋅⋅= ( ) ( )x5ln2lnx652x'f 2x1x3 24 += + OR
  • 11. yx53.8 4yx +=+ ( ) ( ) 'yx4'y5ln53ln3 3yx +=+ ( )[ ] ( )3ln3x415ln5'y x3y −=− ( ) ( )[ ]15ln5 3ln3x4 'y y x3 − − =
  • 12. A. Find the derivative and simplify the result. ( ) 1x3x2 3xg.1 +− = ( ) 22 xlnx exf.2 + = 2e e y.3 x3 x4 + = ( ) 3 x22 2 3xlogxh.4 ⋅= ( ) 2 x 5xG.1 = 2lnyxxeye.2 22yx ++=+ ( ) ( ) 2 X 1xxH.3 += 1x2 e y.4 1x2 + = + ( ) ( )x2x2 eelnxf.5 − += B. Apply the appropriate formulas to obtain the derivative of the given function and simplify. EXERCISES:
  • 13. Logarithmic Differentiation Oftentimes, the derivatives of algebraic functions which appear complicated in form (involving products, quotients and powers) can be found quickly by taking the natural logarithms of both sides and applying the properties of logarithms before differentiation. This method is called logarithmic differentiation.
  • 14. 1. Take the natural logarithm of both sides and apply the properties of logarithms. 2. Differentiate both sides and reduce the right side to a single fraction. 3. Solve for y’ by multiplying the right side by y. 4. Substitute and simplify the result. Steps in applying logarithmic differentiation. Logarithmic differentiation is also applicable whenever the base and its power are both functions.
  • 15. x xyif dx dy Find.1 = xlnxyln xlnyln x = = Logarithmic differentiation is also applicable whenever the base and its power are both functions. (Variable to variable power.) Example: ( ) ( )1xln1 x 1 x'y y 1 += ( ) x xybutyxln1'y =→+= ( )( )x xxln1'y +=∴
  • 16. ( ) ( ) ( )1x2ln1xyln 1x2lnyln 1x +−= += − ( ) 1x 1x2yif dx dy Find.2 − += ( ) ( ) ( )( )11x2ln2 1x2 1 1x'y y 1 ++ + −= ( ) ( )1x2ln 1x2 1x2 'y y 1 ++ + − = ( ) ( ) ( ) 1-x 12xybuty1x2ln 1x2 1x2 'y +=→      ++ + − = ( ) ( ) ( ) ( ) 1-x 12x 1x2 1x2ln1x21x2 'y +      + +++− = ( ) ( ) ( )[ ]( ) 1-1-x 12x1x2ln1x21x2'y ++++−= ( ) ( ) ( )[ ]( ) 2-x 12x1x2ln1x21x2'y ++++−=∴
  • 17. ( ) x 5x6y.3 += ( ) 5x6lnxyln 5x6lnyln x += +=       ++      +       + = x2 1 5x6ln 5x62 6 5x6 1 xy' y 1 ( ) ( ) 1 2 x 5x6 x2 5x6ln5x6x6 y' − +      +++ =∴ x2 5x6ln 5x6 x3 y' y 1 + + + = ( ) ( )5x6x2 5x6ln5x6x6 y' y 1 + +++ = ( ) ( ) ( ) ( ) x 5x6ybuty 5x6x2 5x6ln5x6x6 y' +=⇒      + +++ = ( ) ( ) ( ) x 5x6 5x6x2 5x6ln5x6x6 y' +      + +++ =
  • 18. ( ) 1x x34y.4 − −= ( ) ( )x34ln1xyln x34lnyln 1x −−= −= − ( ) ( )       − −+      − − −= 1x2 1 x34ln3 x34 1 1x'y y 1 ( ) 1x2 x34ln x34 1x3 'y y 1 − − + − −− = ( ) ( ) ( ) ( ) 1xx342 x34lnx341x6 'y y 1 −− −−+−− = ( ) ( ) ( ) ( ) ( ) ( ) 1x x34ybuty 1xx342 x34lnx341x6 y' − −=⇒      −− −−+−− = ( ) ( ) ( ) ( ) ( ) 1x x34 1xx342 x34lnx341x6 'y − −      −− −−+−− = ( ) ( ) ( ) ( ) 11x x34 1x2 x34lnx341x6 y' −− −      − −−+−− =∴