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

As we have just seen in part 1 of Differentiating Log functions, there are a number of different types of questions and approaches.

We continue these now with some more involved questions. 

 
Example 1

$y=\ln\left[\left(\frac{x+1}{x-1}\right)^2\right]$y=ln[(x+1x1)2]

$y$y $=$= $\ln\left[\left(\frac{x+1}{x-1}\right)^2\right]$ln[(x+1x1)2]
$y$y $=$= $2\ln\left(\frac{x+1}{x-1}\right)$2ln(x+1x1)
  $=$= $2\left(\ln\left(x+1\right)-\ln\left(x-1\right)\right)$2(ln(x+1)ln(x1))
$\therefore\frac{dy}{dx}$dydx $=$= $2\left(\frac{1}{x+1}-\frac{1}{x-1}\right)$2(1x+11x1)
  $=$= $\frac{-2}{x^2-1}$2x21
     
Example 2

Find the gradient of the tangent at $x=1$x=1 on the curve given by $y=3\ln\left[\left|2x-1\right|\right]+e^{2x-1}$y=3ln[|2x1|]+e2x1

$y$y $=$= $3\ln\left[\left|2x-1\right|\right]+e^{2x-1}$3ln[|2x1|]+e2x1
$\therefore\frac{dy}{dx}$dydx $=$= $\frac{6}{2x-1}+2e^{2x-1}$62x1+2e2x1
     

Thus, at $x=1$x=1$\frac{dy}{dx}=6+2e$dydx=6+2e. Hence the gradient of the tangent is $6+2e$6+2e.

Example 3

Find the gradient and the point of contact of the tangent to the curve given by $y=\ln\left(\frac{1}{\left(x+1\right)^3}\right)$y=ln(1(x+1)3) at $x=2$x=2.

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

At $x=2$x=2$\frac{dy}{dx}=\frac{-3}{3}=-1$dydx=33=1 and $y=-3\ln\left(2+1\right)=-3\ln3$y=3ln(2+1)=3ln3.

Example 4

Is the function given as $y=x-\ln x$y=xlnx strictly decreasing?

If the function is strictly decreasing, then it is decreasing across the whole of its natural domain. Note that there is a subtle difference between the phrases strictly decreasing and strictly non-increasing.

If,at a certain point, a function is decreasing then the gradient at that point is negative. 

The gradient function is given by $\frac{dy}{dx}=1-\frac{1}{x}$dydx=11x. We can see that at, say $x=3$x=3$\frac{dy}{dx}=\frac{2}{3}$dydx=23, so there is at least one point where the function is increasing. Therefore it is not strictly decreasing.

Example 5

Is the function given as $y=-3\ln\left(x+1\right)$y=3ln(x+1) strictly decreasing?

The natural domain of the function $y=-3\ln\left(x+1\right)$y=3ln(x+1) is determined by solving the inequality $x+1>0$x+1>0 from which we see that $x>-1$x>1.

We can write that $\frac{dy}{dx}=\frac{-3}{x+1}$dydx=3x+1 and look for any value of $x$x within the natural domain for which $\frac{dy}{dx}>0$dydx>0. Arguing logically, if $x>-1$x>1, then the denominator of the derivative, $x+1$x+1, is positive, and thus the derivative itself is always negative. Hence the function is strictly decreasing.

Example 6

Find the equation of the tangent to the curve given by $y=\log_2\sqrt{x}$y=log2x at the point where $x=4$x=4

From the derivative rules above, we determine the derivative as follows:

$y$y $=$= $\log_2\sqrt{x}$log2x
$y$y $=$= $\frac{1}{2}\log_2x$12log2x
$\therefore\frac{dy}{dx}$dydx $=$= $\frac{1}{2x\ln2}$12xln2
     

  At $x=4$x=4$y=\log_2\sqrt{4}=\log_22=1$y=log24=log22=1, and $\frac{dy}{dx}=\frac{1}{8\ln2}$dydx=18ln2

Therefore using the point-gradient form of the equation of a line given by $y-y_1=m\left(x-x_1\right)$yy1=m(xx1), we can write:

$y-1$y1 $=$= $\frac{1}{8\ln2}\left(x-4\right)$18ln2(x4)
$\therefore y$y $=$= $\frac{x}{8\ln2}+1-\frac{1}{2\ln2}$x8ln2+112ln2
     

Example 7

Consider the function $f\left(x\right)=\ln\left(\sqrt{x^2+1}\right)$f(x)=ln(x2+1).

  1. Find $f'\left(x\right)$f(x).

  2. Hence find $f'\left(2\right)$f(2).

 

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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