Question

How to differentiate cube root of sin x with respect to x? Tell me correct way please.

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: \sqrt[3]{sinx}$

Let assume that

$\rm \: y = \sqrt[3]{sinx}$

can be rewritten as

$\rm \: y = {\bigg(sinx\bigg) }^{\dfrac{1}{3} }$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg(sinx\bigg) }^{\dfrac{1}{3} }$

We know,

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: }}$

and

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}f[g(x)] = f'[g(x)] \: \dfrac{d}{dx}g(x) \: }}$

So, using these results, we get

$\rm \: \dfrac{dy}{dx}= \dfrac{1}{3} {\bigg(sinx\bigg) }^{\dfrac{1}{3} \: - \: 1 } \dfrac{d}{dx}sinx$

$\rm \: \dfrac{dy}{dx}= \dfrac{1}{3} {\bigg(sinx\bigg) }^{\dfrac{1 - 3}{3}} \times cosx$

$\rm \: \dfrac{dy}{dx}= \dfrac{1}{3} {\bigg(sinx\bigg) }^{\dfrac{- 2}{3}} \times cosx$

$\rm\implies \:\dfrac{dy}{dx} = \dfrac{cosx}{3 {\bigg(sinx \bigg) }^{\dfrac{2}{3} } }$

Or

$\rm\implies \:\dfrac{dy}{dx} = \dfrac{cosx}{3 \: \sqrt[3]{ {sin}^{2} x} }$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

Let,

$\tt \[ f(x)=\sqrt[3]{\sin x} \]$

By using first principles

$\red{\[ \begin{array}{l} \tt f^{\prime}(x)=\underset{h \rightarrow 0}{ \lim} \dfrac{f(x+h)-f(x)}{h} \\ \\ \tt f^{\prime}(x)=\underset{h \rightarrow 0} { \lim}\dfrac{\sqrt[3]{\sin (x+h)}-\sqrt[3]{\sin x}}{h} \end{array} \]}$

$\text{The above is of \: \( \dfrac{0}{0} \) \: \: form so apply L'Hopital's rule,}$

$\color{darkcyan}\[ \begin{array}{l} \tt =\underset{h \rightarrow 0}{\lim} \dfrac{\dfrac{\cos (x+h)}{3 \sin ^{\frac{2}{3}}(x+h)}-0}{1} \qquad\left[\text { Since } \dfrac{d}{d x}\right. \left.(\sqrt[3]{\sin (x)})=\dfrac{\cos (x)}{3 \sin ^{\frac{2}{3}}(x)}\right] \end{array} \]$

Apply the limit h=0.

$\color{olive}\[ \begin{array}{l} \tt f^{\prime}(x)=\dfrac{\cos x}{3 \sin ^{\frac{2}{3}} x} \\ \\ \tt \Longrightarrow f^{\prime}(x)=\dfrac{\cos x}{3 \sqrt[3]{\sin ^{2} x}} \end{array} \]$