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Mathematics Learning Centre

Derivatives of exponential and

logarithmic functions

Christopher Thomas

c?1997 University of Sydney Mathematics Learning Centre, University of Sydney1

1Derivatives of exponential and logarithmic func-

tions If you are not familiar with exponential and logarithmic functions you may wish to consult the bookletExponents and Logarithmswhich is available from the Mathematics Learning

Centre.

Youmayhave seen that there are two notations popularly used for natural logarithms, log e and ln. These are just two different ways of writing exactly the same thing, so that log e x≡lnx.Inthis booklet we will use both these notations.

The basic results are:

d dxe x =e x d dx(log e x)=1 x. Wecan use these results and the rules that we have learnt already to differentiate functions which involve exponentials or logarithms.

Example

Differentiate log

e (x 2 +3x+1).

Solution

Wesolve this by using the chain rule and our knowledge of the derivative of log e x. d dxlog e (x 2 +3x+1) =d dx(log e u)(whereu=x 2 +3x+1) d du(log e u)×du dx(by the chain rule) 1 u×dudx 1 x 2 +3x+1×ddx(x 2 +3x+1) 1 x 2 +3x+1×(2x+3) 2x+3 x 2 +3x+1.

Example

Find d dx (e 3x 2 Mathematics Learning Centre, University of Sydney2

Solution

This is an application of the chain rule together with our knowledge of the derivative of e x d dx(e 3x 2 )=de u dxwhereu=3x 2 =de u du×dudxbythe chain rule =e u

×du

dx =e 3x 2 ×d dx(3x 2 =6xe 3x 2

Example

Find d dx (e x 3 +2x

Solution

Again, we use our knowledge of the derivative ofe

x together with the chain rule. d dx(e x 3 +2x )=de u dx(whereu=x 3 +2x) =e u

×du

dx(by the chain rule) =e x 3 +2x ×d dx(x 3 +2x) =(3x 2 +2)×e x 3 +2x

Example

Differentiate ln(2x

3 +5x 2 -3).

Solution

Wesolve this by using the chain rule and our knowledge of the derivative of lnx. d dxln(2x 3 +5x 2 -3) =dlnu dx(whereu=(2x 3 +5x 2 -3) dlnu du×dudx(by the chain rule) 1 u×dudx 1 2x 3 +5x 2 -3×ddx(2x 3 +5x 2 -3) 1 2x 3 +5x 2 -3×(6x 2 +10x) 6x 2 +10x 2x 3 +5x 2 -3. Mathematics Learning Centre, University of Sydney3 There are two shortcuts to differentiating functions involving exponents and logarithms.

The four examples above gave

d dx(log e (x 2 +3x+1)) =2x+3 x 2 +3x+1 d dx(e 3x 2 )=6xe 3x 2 d dx(e x 3 +2x )=(3x 2 +2)e 3x 2 d dx(log e (2x 3 +5x 2 -3)) =6x 2 +10x 2x 3 +5x 2 -3.

These examples suggest the general rules

d dx(e f(x) )=f (x)e f(x) d dx(lnf(x)) =f (x) f(x). These rules arise from the chain rule and the fact that de x dx =e x and dlnx dx 1 x .They can speed up the process of differentiation but it is not necessary that you remember them. If you forget, just use the chain rule as in the examples above.

Exercise 1

Differentiate the following functions.

a.f(x)=ln(2x 3 )b.f(x)=e x 7 c.f(x)=ln(11x 7 d.f(x)=e x 2 +x 3 e.f(x)=log e (7x -2 )f.f(x)=e -x g.f(x)=ln(e x +x 3 )h.f(x)=ln(e x x 3 )i.f(x)=ln x 2 +1 x 3 -x Mathematics Learning Centre, University of Sydney4

Solutions to Exercise 1

a.f (x)=6x 2 2x 3 =3 x

Alternatively writef(x)=ln2+3lnxso thatf

(x)=31 x. b.f (x)=7x 6 e x 7 c.f (x)= 7 x d.f (x)=(2x+3x 2 )e x 2 +x 3 e.Writef(x)=log e

7-2log

e xso thatf (x)=- 2 x f.f (x)=-e -x g.f (x)=e x +3x 2 e x +x 3 h.Writef(x)=lne x 3 lnx so thatf (x)=1+3 x. i.Writef(x)=ln(x 2 +1)-ln(x 3 -x)sothatf (x)=2x x 2 +1-3x 2 -1 x 3 -x.quotesdbs_dbs12.pdfusesText_18
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