# Thread: Help with finding the derivative of inverse functions

1. ## Help with finding the derivative of inverse functions

Hey I need some help with the following problems:

Find (f^-1)'(a):

1. f(x)= 2x^3 + 3x^2 + 7x + 4, a=4

2. f(x)= x^3 + 3sinx + 2cosx, a=2

2. If $\mathrm{f}^{-1}(a)=b$, then $a=\mathrm{f}(b)$. Find b from the given a.

Now use this rule: If $y=\mathrm{f}(x)$, $(\mathrm{f}^{-1})'(y)=\frac{1}{\mathrm{f}'(x)}$. Hence $(\mathrm{f}^{-1})'(a)=\frac{1}{\mathrm{f}'(b)}$.

For example, for (1): $\mathrm{f}(x)=4\$ $\Rightarrow\$ $2x^3 + 3x^2 + 7x + 4=4$ $\Rightarrow\$ $x(2x^2+3x+7)=0$ $\Rightarrow\$ $x=0$ (the quadratic has no real roots). So $(\mathrm{f}^{-1})'(4)=\frac{1}{\mathrm{f}'(0)}=\frac{1}{7}$

Similarly for (2). $\mathrm{f}(x)=2\$ $\Rightarrow\$ $x^3 + 3\sin{x} + 2\cos{x}=2$. By inspection, $x=0$ is a solution. Hence $(\mathrm{f}^{-1})'(2)=\frac{1}{\mathrm{f}'(0)}$.
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