1. ## inverse function differentiation

Let
$\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)

2. Originally Posted by 234578
Let
$\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)
Dear 234578,

Try integrating the expression and obtain F(x) using the substitution, $\displaystyle t=tan\theta$

Afterwards you could find the inverse of F(x) and G'(0)

3. Originally Posted by 234578
Let
$\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)
since F and G are inverses ...

$\displaystyle F[G(x)] = x$

$\displaystyle F'[G(x)] \cdot G'(x) = 1$

$\displaystyle G'(x) = \frac{1}{F'[G(x)]}$

$\displaystyle G'(0) = \frac{1}{F'[G(0)]}$

since $\displaystyle F(1) = 0$ , $\displaystyle G(0) = 1$

$\displaystyle G'(0) = \frac{1}{F'(1)}$