inverse function derivative

**gracy** · February 4th 2007, 07:25 AM

what is the value of (d/dx)[f^(-1) (x) ] when x=-2 given that f(x)=x^3+x.

**CaptainBlack** · February 4th 2007, 09:21 AM

Originally Posted by gracy

what is the value of (d/dx)[f^(-1) (x) ] when x=-2 given that f(x)=x^3+x.

First of all f(x) is an increasing function of x so the inverse function exists on $\mathbb{R}$

Let $f(y)=x$ , then

$\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}=(3y^2+1)\frac{ dy}{dx}=1$ .

Hence:

$\frac{dy}{dx}=\frac{1}{3y^2+1}=\frac{1}{3\,f(x)+1} =\frac{1}{3x^3+3x+1}$

but $y=f^{-1}(x)$ , so:

$\frac{d}{dx}f^{-1}(x)=\frac{1}{3x^3+3x+1}$ .

So finally:

$\left. \frac{d}{dx}f^{-1}(x)\right|_{x=-2}=\frac{1}{-3\times 8-3 \times 2+1}\approx -0.0345$ .

RonL

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