Question

find dy/dx of 1/x plz with explain I will give brain mark plz plz ​

Answer 1

$\underline{\mathrm{Solution :}}$

$\boxed{\star}\boxed{\textsf{Direct method-}}\boxed{\star}$

$\mathrm{Let,\:y=\frac{1}{x}}$

$\textsf{Diff. both sides w. r. to x, we get}$

$\mathrm{\frac{dy}{dx}=\frac{d}{dx}(\frac{1}{x})}$

$\mathrm{=\frac{d}{dx}(x^{-1})}$

$\mathrm{=(-1) * x^{-1-1}}$

$\mathrm{=-x^{-2}}$

$\mathrm{=-\frac{1}{x^{2}}}$

$\implies \boxed{\:\:\mathrm{\frac{dy}{dx}=-\frac{1}{x^{2}}}\:\:}$

$\underline{\textsf{Formula :}}$

$\mathrm{\frac{d}{dx}(x^{n})=n\:x^{n-1}\:,\:n\in \mathbb{R}}$

$\boxed{\star}\boxed{\textsf{Using first principle-}}\boxed{\star}$

$\mathrm{Let,\:f(x)=\frac{1}{x}}$

$\mathrm{Then,\:\frac{dy}{dx}}$

$\mathrm{=lim_{h \to 0} \frac{f(x+h)-f(x)}{h}}$

$\mathrm{=lim_{h \to 0} \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}}$

$\mathrm{=lim_{h \to 0} \dfrac{\frac{x-x-h}{(x+h)x}}{h}}$

$\mathrm{=lim_{h \to 0} \frac{-h}{xh(x+h)}}$

$\mathrm{=lim_{h \to 0} \frac{-1}{x(x+h)}}$

$\mathrm{=-\frac{1}{x * x}}$

$\mathrm{=-\frac{1}{x^{2}}}$

$\to \boxed{\mathrm{\frac{dy}{dx} = -\frac{1}{x^{2}}}}$