Question

differenciate w.r.t to x $y=\dfrac{2}{x}+\frac{x}{2}$ Ans:$-2x^{-2}+\frac{1}{2}$​

Answer 1

$\huge\bf{\red{\overbrace{\underbrace{\purple{Given:}}}}}$

★$y=\dfrac{2}{x}+\frac{x}{2}$

$\huge\bf{\red{\overbrace{\underbrace{\purple{To\:\:Find:}}}}}$

★The differentiation of y w. r. t. x.

$\huge\bf{\red {\overbrace{\underbrace{\purple{Answer:}}}}}$

We have,

$y=\dfrac{2}{x}+\frac{x}{2}$

So,

$\implies \dfrac{dy}{dx} =\dfrac{d[\dfrac{2}{x}+\dfrac{x}{2}]}{dx}$

$\implies \dfrac{dy}{dx}=\dfrac{d[2x^{-1}+\frac{1}{2}(x) ]}{dx}$

$\implies \dfrac{dy}{dx}=\dfrac{d[2x^{-1}]}{dx}+\dfrac{1}{2}\dfrac{d[x]}{dx}$

$\implies \dfrac{dy}{dx}=2\frac{d[x^{-1}]}{dx}+\dfrac{1}{2}\times 1$

$\large\orange{\boxed{\bf{\purple{y=a^{n}, \dfrac{dy}{dx}=na^{n-1}}}}}$

$\implies \dfrac{dy}{dx}=-2x^{-2}+\dfrac{1}{2}$

$\large\green{\boxed{\red{\sf{.\degree.\dfrac{dy}{dx}=-2x^{-2}+\dfrac{1}{2}}}}}$

Answer 2

$\frac{dy}{dx} = { - 2x}^{ - 2} + \frac{1}{2}$

Explanation:

Given,

$y = \frac{2}{x} + \frac{x}{2}$

$\frac{dy}{dx} = \frac{d{ \frac{2}{x} + \frac{x}{2}} }{dx}$

$\frac{dy}{dx} = \frac{d( {2x}^{ - 1 + \frac{1}{2}x) } }{dx}$

$\frac{dy}{dx} = \frac{d( {2x}^{ - 1)} }{dx} + \frac{1}{2} \frac{d(x)}{dx}$

$\frac{dy}{dx} = 2 \frac{d( {x}^{ - 1)} }{dx} + \frac{1}{2} \times 1$

$y = {a}^{n}$

$\frac{dy}{dx} = {na}^{n - 1}$

so,

The answer is

$\frac{dy}{dx} = { - 2x}^{ - 2} + \frac{1}{2}$