Question

Find a function $f$that has the given derivative $f^{\text{'}}$and value f(c)

$f^{\text{'}}\left(x\right)=2xsec{x}^2\tan x^2,f\left(0\right)=2$

Short answer

The anti derivative is$sec\left(x^2\right)+2$

Step-by-step solution

Given information 

The given expression is$f^{\text{'}}\left(x\right)=2xsec{x}^2\tan x^2,f\left(0\right)=2$

Simplification

Given $f^{\text{'}}\left(x\right)=2xsec{x}^2\tan x^2,f\left(0\right)=2$

We will use an intelligent guess-and-check method to find f

Here $sec{x}^2\tan x^2$is a derivative of $sec{x}^2$

So our guess can be $sec{x}^2$

Now, $2x$is a derivative of $x^2$

Hence we can say that

$f\left(x\right)=sec{x}^2$$+c$

Now,

$f\left(0\right)=2$

$\Rightarrow 2=sec\left(0\right)+c$

localid="1660720299977" $$\begin{gathered} \Rightarrow 2=1+c \\ \Rightarrow c=2-1 \\ \Rightarrow c=1 \end{gathered}$$

Hence the antiderivative is $se{c}^{2}x+1$