NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Differentiating Logarithmic functions I
Lesson

Armed with the general derivatives of log and exponential functions, and making use of the product, quotient and chain rules and the working rules of logs, we can now find derivatives for a large group of functions.

Let's quickly review these specific results:

With $u$u and $v$v as functions of $x$x, we can write:

Table A: Working rules of Logs

Expression Equivalence
$\log_b\left(uv\right)$logb(uv) $\log_bu+\log_bv$logbu+logbv
$\log_b\left(\frac{u}{v}\right)$logb(uv) $\log_bu-\log_bv$logbulogbv
$\log_bu^n$logbun $n\log_bu$nlogbu

 

Table B: Derivatives

 
Concept Function Derivative
Exponential $y=e^u$y=eu $\frac{dy}{dx}=u'\times e^u$dydx=u×eu
Natural log $y=\ln u$y=lnu $\frac{dy}{dx}=\frac{u'}{u}$dydx=uu
Any log $y=\log_bu$y=logbu $\frac{dy}{dx}=\frac{u'}{u\times\ln b}$dydx=uu×lnb
Product $y=uv$y=uv $\frac{dy}{dx}=uv'+vu'$dydx=uv+vu
Quotient $y=\frac{u}{v}$y=uv $\frac{dy}{dx}=\frac{vu'-uv'}{v^2}$dydx=vuuvv2
Chain $y=u^n$y=un $\frac{dy}{dx}=nu^{n-1}\times u'$dydx=nun1×u

The following examples illustrate how this knowledge is applied. You will note that in many instances, expressions involving logs can be simplified (using the working rules of logs) before differentiating. 

Example 1 

$y=\ln\left(2x^3-x+1\right)$y=ln(2x3x+1)

$y$y $=$= $\ln\left(2x^3-x+1\right)$ln(2x3x+1)
$\therefore\frac{dy}{dx}$dydx $=$= $\frac{6x^2-1}{2x^3-x+1}$6x212x3x+1
     

 

Example 2

$y=x^2\ln2x$y=x2ln2x

$y$y $=$= $x^2\ln2x$x2ln2x
$\therefore\frac{dy}{dx}$dydx $=$= $x^2\left(\frac{2}{2x}\right)+2x\ln2x$x2(22x)+2xln2x
  $=$= $x+2x\ln2x$x+2xln2x
  $=$= $x\left(1+2\ln2x\right)$x(1+2ln2x)
     
Example 3

$y=\frac{\ln\left(x+1\right)}{e^x}$y=ln(x+1)ex

$y$y $=$= $\frac{\ln\left(x+1\right)}{e^x}$ln(x+1)ex
$\therefore\frac{dy}{dx}$dydx $=$= $\frac{e^x\left(\frac{1}{x+1}\right)-e^x\ln\left(x+1\right)}{\left(e^x\right)^2}$ex(1x+1)exln(x+1)(ex)2
  $=$= $\frac{\left(\left(\frac{1}{x+1}\right)-\ln\left(x+1\right)\right)}{e^x}$((1x+1)ln(x+1))ex
  $=$= $\frac{1-\left(x+1\right)\ln\left(x+1\right)}{e^x\left(x+1\right)}$1(x+1)ln(x+1)ex(x+1)
     
Example 4

$y=\ln\sqrt{3x^2-1}$y=ln3x21

$y$y $=$= $\ln\sqrt{3x^2-1}$ln3x21
$y$y $=$= $\ln\left(3x^2-1\right)^{\frac{1}{2}}$ln(3x21)12
$y$y $=$= $\frac{1}{2}\ln\left(3x^2+1\right)$12ln(3x2+1)
$\therefore\frac{dy}{dx}$dydx  $=$= $\frac{1}{2}\times\frac{6x}{3x^2+1}$12×6x3x2+1
  $=$= $\frac{3x}{3x^2+1}$3x3x2+1
     

Some Mathspace Examples

Question 1

Find the derivative of $y=\ln\left(x^4+2\right)$y=ln(x4+2). You may use the substitution $u=x^4+2$u=x4+2.

Question 2

Find the derivative of $y=6x^4\ln x$y=6x4lnx.

Question 3

 

Consider the function $y=\ln\left(\frac{x-3}{x+3}\right)$y=ln(x3x+3).

  1. Let $u=\frac{x-3}{x+3}$u=x3x+3.

    Rewrite $y$y in terms of $u$u.

  2. Determine $\frac{du}{dx}$dudx. Leave the denominator in factorised form.

  3. Next determine $\frac{dy}{du}$dydu. Express your answer in terms of $x$x.

  4. Hence, or otherwise, determine $\frac{dy}{dx}$dydx.

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91578

Apply differentiation methods in solving problems

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