I am working from a textbook that gives us that
The derivative of the inverse function is $$ \frac{dx}{dy} = \large{\frac{1}{\frac{dy}{dx}}}.\qquad (*)$$
$$$$Now I am stuck on part $(a)$ of the following question:
A function is defined by $\ f(x) =x^3 + 3x + 2.$
$(a)$ By considering $f'(x),\ $ prove that $\ f(x)\ $ has an inverse function.
$(b)\ $ Find the gradient of the graph of $y=f^{-1}(x)$ at the point where $x=2.$
$$$$For $(b),\ f(x) = 2\ \iff x=0\ $. Therefore, the gradient of the graph of $y=f^{-1}(x)$ at the point where $\ x=2\ $ is equal to $\ \large{\frac{1}{f'(0)}} = \frac{1}{3}.$
But I'm simply a bit confused unsure on what it wants for part $(a)$... The derivative of the inverse function is $\ \large{\frac{1}{3}\cdot \frac{1}{x^2+1} },\ $ but this does not say that, for example, the derivative of the inverse function $\ y=f^{-1}(x)\ $ at the point $\ x=3\ $ is $\ \large{\frac{1}{3}\cdot \frac{1}{3^2+1} },\ $ which it isn't. This means I find $(*)$ confusing as to what it is saying exactly, and like I say, I'm confused about this and not sure how to tackle part $(a)$. Please can someone clarify. Thanks.
$\endgroup$3 Answers
$\begingroup$The derivative of an inverse can be written more clearly as$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))},$$and part $(a)$ can be shown straightforwardly using the inverse function theorem ().
$\endgroup$ $\begingroup$(a) Let g(.) be the inverse function of f(.)
f'(x) = 3(x^2 + 1) > 0 for all x.
f is strictly increasing, therefore, it is one to one.
Hence its inverse g: R-->R, defined by g(x) = $f^{-1}(x)$and g'(f(x)) = 1/f'(x) By definition g(f(x)) = x. Differentiating wrt x, by chain rule of differentiation, we have g'(f(x))f'(x)= 1 or g'(f(x)) = 1/f'(x).(b) g'(f(2)) = $\frac{1} {3(2^2 + 1)}$ = $\frac{1} {15}$
$\endgroup$ 5 $\begingroup$$(a)\quad f'(x) =3(x^2 +1)\geq3>0\ $ for all $\ x.\ $ Therefore $\ f(x)\ $ is strictly increasing, and therefore one-to-one. Since a one-to-one function has an inverse, $\ f(x)\ $ has an inverse, i.e. $\ f^{-1}(x)\ $ exists.
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