Question

If $\frac{d}{dx}(f(x\left)\right)=\frac{1}{(1+x^2)}$, then $\frac{d}{dx}(f(x^3\left)\right)=$

• A. $\frac{3x}{(1+x^3)}$
• B. $\frac{3x^2}{(1+x^6)}$
• C. $-\frac{6x^5}{{(1+x^6)}^2}$
• D. $-\frac{6x^5}{(1+x^6)}$

Answer 1

The correct option is B

$\frac{3x^2}{(1+x^6)}$

Find the value of $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathit{f}\mathbf{}\mathbf{(}{\mathit{x}}^{\mathbf{3}}\mathbf{\left)}\mathbf{\right)}$:

Given, $\frac{d}{dx}(f(x\left)\right)=\frac{1}{(1+x^2)}$

$\Rightarrow$$\int \frac{d}{dx}(f(x\left)\right)dx=\int \frac{1}{(1+x^2)}dx$

$\Rightarrow$ $f\left(x\right)={\tan }^{-1}x+C$

$\Rightarrow$ $f\left(x^3\right)={\tan }^{-1}\left(x^3\right)+C$

$\therefore \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{f}\mathbf{\left(}{\mathbf{x}}^{\mathbf{3}}\mathbf{\right)}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}^{\mathbf{6}}}$

Hence, Option ‘B’ is Correct.