Question

Find an equation of the tangent line to the graph of the function at the given point.

Answer 1

We will have the following:

First, we determine the derivative of the expression, that is:

$y=9ln(\frac{e^x+e^{-x}}{2})\Rightarrow y^{\prime}=\frac{9(e^{2x}-1)}{1+e^{2x}}$

Now, we determine the value of the slope at x = 0, that is:

$y^{\prime}(0)=\frac{9(e^{2(0)}-1)}{1+e^{2(0)}}\Rightarrow y^{\prime}(0)=\frac{9(1-1)}{1+1}\Rightarrow y^{\prime}(0)=0$

So, from this we will have that the equation of the line that is tangent for the function at the point (0, 0) will be:

$y-0=0(x-0)\Rightarrow y=0$

So, the equation of the line will be:

$y=0$

This can be seeing as follows: