Question

Consider the following function. Without finding the​ inverse, evaluate the derivative of the inverse at the given point. f(x)=ln(8x+e); (1,0)

Answer 1

We can use the inverse function derivative theorem:

$\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.$

In this case, we want to evaluate $\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}$, so:

$\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.$

The derivative is:

$\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.$

The ordinate of the point is $f^{-1}(1) = 0$, so we evaluate:

$\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.$

Finally:

$\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.$

We can check the answer by finding the inverse:

$y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},$

so that

$f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.$

Therefore:

$\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.$

Which finally gives the same answer as before:

$\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.$

Answer: $\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.$