# Let f(x)=\int_{6}^{x}\frac{dt}{\sqrt{3+t^{4}}}. Find {(f^{-1})}'(0)..

## Question:

Let {eq}f(x)=\int_{6}^{x}\frac{dt}{\sqrt{3+t^{4}}} {/eq}.

Find {eq}{(f^{-1})}'(0). {/eq}

## Derivative Of An Inverse Function:

To compute the derivative of an inverse function at a given value of x, we use the formula:

{eq}\left(f^{-1}\right)^{\prime}(a)=\dfrac{1}{f^{\prime}\left(f^{-1}(a)\right)} {/eq}

Note: If the function is defined as a definite integral whose lower limit is a number and the upper digit is a variable, then we use the fundamental theorem of calculus to find its derivative which states:

$$\dfrac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

## Answer and Explanation:

The given function is:

$$f(x)=\int_{6}^{x} \frac{d t}{\sqrt{3+t^{4}}}$$

The derivative of this function is found using the fundamental theorem of calculus which states:

$$\dfrac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

Then we get:

$$f'(x)= \dfrac{d}{dx}\int_{6}^{x} \frac{d t}{\sqrt{3+t^{4}}}= \frac{1}{\sqrt{3+x^{4}}}$$

The given value of 'c' is:

$$c=0$$

First, we will find {eq}f^{-1}(0) {/eq}.

Let us assume that:

$$f(x) = 0 \\ \int_{6}^{x} \frac{d t}{\sqrt{3+t^{4}}}=0$$

Clearly, {eq}x=6 {/eq} satisfies this (because the value of a definite integral is zero when its bounds are same).

So we get:

$$f(6)=0 \Rightarrow f^{-1}(0)=6 \,\,\, \rightarrow (1)$$

Substitute {eq}c=0 {/eq} in the formula:

\begin{align} \left(f^{-1}\right)^{\prime}(c)&=\dfrac{1}{f^{\prime}\left(f^{-1}(c)\right)} \\ \left(f^{-1}\right)^{\prime}(0)&=\dfrac{1}{f^{\prime}\left(f^{-1}(0)\right)} \\ &= \dfrac{1}{f'(6)} & \text{(From (1))} \\ & = \dfrac{1}{\frac{1}{\sqrt{3+6^{4}}}} \\ & = \sqrt{1299} \end{align}

Therefore: {eq}\boxed{\mathbf{\left(f^{-1}\right)^{\prime}(0) = \sqrt{1299}}} {/eq} 