Question

$\huge{\underline{\underline{\red{\mathfrak{Question :}}}}}$

Derivative of the function using first principle :

³√Sinx ​

Answer 1

$\tt f(x) = \sqrt[3]{\sin \tt x} \\ \\ \implies f(x+h) = \sqrt[3]{\sin\tt (x+h) } \\ \\ We\ know\ that\ :\\ \\ \dfrac{d}{dx}\big(f(x)\big) =  \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ \\ \\ \implies \dfrac{\sqrt[3]{\sin\tt (x+h)}-\sqrt[3]{\sin\tt (x) } }{h} \\ \\ We\ know\ that: a-b=\dfrac{a^3-b^3}{a^2+b^2+ab}$

$\tt \implies  \lim_{h \to 0} \dfrac{\sqrt[3]{\sin\tt (x+h)}-\sqrt[3]{\sin\tt (x) } }{h}\\ \\ \implies \lim_{h \to 0} \dfrac{\Big(\sqrt[3]{\sin\tt (x+h)}\Big)^3-\Big(\sqrt[3]{\sin\tt (x) }\Big)^3 }{h\times \Big(\sin^\frac{2}{3}\tt(x+h)+\sin^\frac{1}{3}\tt(x)\cdot\sin^\frac{1}{3}\tt(x+h)+\sin^\frac{2}{3}\tt(x)\Big)}$

$\tt \lim_{h \to 0} \dfrac{\sin\tt(x+h)-\sin\tt x}{h} \times \dfrac{1 }{h\times \Big(\sin^\frac{2}{3}\tt(x+h)+\sin^\frac{1}{3}\tt(x)\cdot\sin^\frac{1}{3}\tt(x+h)+\sin^\frac{2}{3}\tt(x)\Big)} \\ \\ \implies  \lim_{h \to 0} \dfrac{2\cos\tt \Big(x+\dfrac{h}{2}\Big)}{2}\times \dfrac{1 }{h\times \Big(\sin^\frac{2}{3}\tt(x+h)+\sin^\frac{1}{3}\tt(x)\cdot\sin^\frac{1}{3}\tt(x+h)+\sin^\frac{2}{3}\tt(x)\Big)}$

$\implies \tt \cos x \times \dfrac{1 }{h\times \Big(\sin^\frac{2}{3}\tt(x)+\sin^\frac{1}{3}\tt(x)\cdot\sin^\frac{1}{3}\tt(x)+\sin^\frac{2}{3}\tt(x)\Big)} \\ \\ \implies \boxed{\tt\dfrac{\cos x}{3.sin^\frac{2}{3}x}}$

Hope this answer helps.

Answer 2

❏ Question :-

Find the derivative of ³√Sinx

❏ Answer :-

$\to \boxed{\sf \frac{dy}{dx} = \frac{ \cos x}{3 \: {( \sin x)}^{ \frac{2}{3} } } } \:$

❏ Solution :-

Let ,

$\to \sf \: y = \sqrt[ 3]{ \sin x} \\ \\ \bf \: we \: can \: write \: it \: \\ \\ \to \sf \: y \: = {( \sin x)}^{ \frac{1}{3} } \\ \\ \bf \: taking \: \log \: on \: both \: sides \: \\ \\ \to \sf \log y = \log {( \sin x)}^{ \frac{1}{3} } \\ \\ \sf \because \log {m}^{n} = n \log m \\ \\ \therefore \\ \to \sf \log y = \frac{1}{3} \log \sin x \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \\ \to \sf \frac{1}{y} \: \frac{dy}{dx} = \frac{1}{3} \frac{1}{ \sin x} \: \frac{d}{dx} \sin x \\ \\ \to \sf \frac{1}{y} \: \frac{dy}{dx} = \frac{ \cos x}{3 \sin x} \\ \\ \to \sf \frac{dy}{dx} = y \bigg( \frac{ \cos x}{3 \sin x} \bigg) \\ \\ \because \sf \: y = {( \sin x)}^{ \frac{1}{3} } \\ \therefore \: \\ \: \to \sf \frac{dy}{dx} = {( \sin x)}^{ \frac{1}{3} } \bigg( \frac{ \cos x}{3 \sin x} \bigg) \\ \\ \to \sf \frac{dy}{dx} = \frac{ \cos x}{3 \: { \sin x}^{(1 - \frac{1}{3} )} } \\ \: \\ \to \boxed{\sf \frac{dy}{dx} = \frac{ \cos x}{3 \: {( \sin x)}^{ \frac{2}{3} } } }$

Another method :-

Let ,

$\to \sf \: y = \sqrt[3]{ \sin x} \\ \\ \: \to \sf y = {( \sin x)}^{ \frac{1}{3} } \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \: \\ \to \sf \frac{dy}{dx} = \frac{1}{3} { \sin x}^{ (\frac{1}{3} - 1)} \: \frac{d}{dx} \sin x \\ \\ \to \sf \frac{dy}{dx} = \frac{1}{3} { \sin}^{( - \frac{2}{3} ) } \: \cos x \\ \\ \to \boxed{ \sf \frac{dy}{dx} = \frac{ \cos x}{3 { \sin}^{ (\frac{2}{3}) }} }$

Hope it help you .