1. The document discusses differentiating exponential functions by applying properties of exponents and logarithms. It provides formulas for differentiating exponentials and natural logarithms. 2. Examples are given of differentiating various exponential functions using the formulas and properties provided. Logarithmic differentiation is also described as a method to differentiate complicated algebraic functions. 3. Steps in applying logarithmic differentiation are outlined, including taking the logarithm of both sides and applying logarithm properties before differentiating.

1. DIFFERENTIATION 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.

3. The EXPONENTIAL FUNCTION
.ylogx
aswrittenbealsomayayfunction,clogarithmi
ofinversetheisfunctionlexponentiatheSince
number.realaisxwhereaybydefinedis1,a
and0aa,basewithfunctionlexponentiaThe
a
x
x
=
=
=≠
>
.

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'
−−
−





−
−−+−−
=∴