Question

Differentiate the function given below w.r.t. $x:$

$y={(x+\frac{1}{x})}^3$

Answer 1

Let $y={(x+\frac{1}{x})}^3$

$\Rightarrow y=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{d\left(x^3\right)}{dx}+\frac{d\left(x^{-3}\right)}{dx}+\frac{d\left(3x\right)}{dx}+\frac{d\left(3x^{-1}\right)}{dx}$

$[\because \frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx}]$

$\Rightarrow \frac{dy}{dx}=3x^2-3x^{-4}+3-3x^{-2}$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$

$\frac{dy}{dx}=3x^2-\frac{3}{x^4}+3-\frac{3}{x^2}$