Question

Find the derivative of $\sqrt[3]{3\sin x}$ from first principles.

Answer 1

$y={\left(\sin x\right)}^{\frac{1}{3}}$

According to first principle , the derivative of $y=f\left(x\right)$

$f^{{}^{\prime }}\left(x\right)=\frac{dy}{dx}={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

Now for $y={\left(\sin x\right)}^{\frac{1}{3}}=f\left(x\right)$

$f(x+h)={\left(\sin (x+h)\right)}^{\frac{1}{3}}\therefore f^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{{\left[\sin (x+h)\right]}^{\frac{1}{3}}-{\left[\sin x\right]}^{\frac{1}{3}}}{h}$

We know that:-

$a^3-b^3=(a-b)(a^2+b^2+ab)\Rightarrow (a-b)=\frac{(a^3-b^3)}{(a^2+b^2+ab)}\therefore f^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{{\left({\left[\sin (x+h)\right]}^{\frac{1}{3}}\right)}^3-{\left({\left(\sin x\right)}^{\frac{1}{3}}\right)}^3}{h[{\sin }^{\frac{2}{3}}(x+h)+{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{1}{3}}(x+h){\sin }^{\frac{1}{3}}x]}={lim}_{h\to 0}\frac{\sin (x+h)-\sin x}{h({\sin }^{\frac{2}{3}}(x+h)+{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{1}{3}}(x+h){\sin }^{\frac{1}{3}}x)}={lim}_{h\to 0}\frac{2\sin \frac{h}{2}\cos (x+\frac{h}{2})}{h({\sin }^{\frac{2}{3}}(x+h)+{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{1}{3}}(x+h){\sin }^{\frac{1}{3}}x)}={lim}_{h\to 0}\frac{\sin \frac{h}{2}}{\frac{h}{2}}{lim}_{h\to 0}\frac{\cos (x+\frac{h}{2})}{{\sin }^{\frac{2}{3}}(x+h)+{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{1}{3}}(x+h){\sin }^{\frac{1}{3}}x}=1\times \frac{\cos x}{{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{2}{3}}x+{\sin }^{\frac{1}{3}}x{\sin }^{\frac{1}{3}}x}=\frac{\cos x}{3{\sin }^{\frac{2}{3}}x}$