Q 32 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 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,

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

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

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

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