Question

Find the function from its given derivative

$f^{\mathrm{\prime }}\left(x\right)=\frac{2x}{\sqrt{1-4x^2}}$

Short answer

The function from its given derivative is

$f\left(x\right)=-\frac{1}{2}\sqrt{1-4x^2}$

Step-by-step solution

Step 1.Given information

WE have been given the derivative of function

$f^{\mathrm{\prime }}\left(x\right)=\frac{2x}{\sqrt{1-4x^2}}$

Step 2.We have been finding the function whose derivative is given 

One can suspect the${\sin }^{-1}x$from the denominator of function

$\begin{array}{r}f\left(x\right)={\sin }^{-1}\left(2x\right)\\ f\left(x\right)={\sin }^{-1}\left(2x\right)\\ f^{\mathrm{\prime }}\left(x\right)=\frac{1}{\sqrt{1-(2x{)}^2}}\frac{d}{dx}\left(2x\right)\\ =\frac{2}{\sqrt{1-4x^2}}\end{array}$

Step 3.Finding the more appropriate function from previous one 

Now , letting the function$f\left(x\right)=-\frac{1}{2}\sqrt{1-4x^2}$

$\begin{array}{r}f\left(x\right)=-\frac{1}{2}\sqrt{1-4x^2}\\ f^{\mathrm{\prime }}\left(x\right)=-\frac{1}{2}\frac{d}{dx}{\left(1-4x^2\right)}^{\frac{1}{2}}\\ =-\left(\frac{1}{2}\right)\frac{1}{2}{\left(1-4x^2\right)}^{-\frac{1}{2}}\frac{d}{dx}\left(1-4x^2\right)\left[\frac{d}{dx}{x}^n=n{x}^{n-1}\right]\\ =-\frac{1}{4\sqrt{1-4x^2}}\left(\frac{d}{dx}1-\frac{d}{dx}4x^2\right)\\ =\frac{1}{A\sqrt{1-4x^2}}\left(A\right(2x\left)\right)\\ =-\frac{-2x}{\sqrt{1-4x^2}}\\ =\frac{2x}{\sqrt{1-4x^2}}\end{array}$