Question

Find the derivative of cube root of (sin x) using first principle?

Answer 1

Hope it will help you

Answer 2

$\frac {1}{3}cotx \sqrt[3]{sinx}$ is the Answer

Step-by-step explanation:

$f(x) = \sqrt[3]{sin x}; f(x^{th}) = \sqrt[3]{sin(x^{th})}$

$f'(x) = \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}$

= $\lim_{h\rightarrow 0}\frac {\sqrt[3]{sin x+h}-\sqrt[3]{sin x}}{h}$

= $\lim_{h\rightarrow 0}[\frac {\sqrt[3]{sin x+h}-\sqrt[3]{sin x}][(\sqrt[3]{sin x+h})^2+\sqrt[3]{sin (x+h)sinx}-(\sqrt[3]{sin x})^2]}{h[\sqrt[3]{sin x+h})^2+\sqrt[3]{sin (x+h)sinx}-(\sqrt[3]{sin x})^2]}$

= $\lim_{h\rightarrow 0}\frac {sin(x+h)-sinx}{h} \times \frac {1}{(sin(x+h))^{\frac {2}{3}}+[sin(x+h)sinx]^{\frac {1}{3}}+sinx^{\frac {2}{3}}}$

= $cos \times \frac {1}{(sinx)^{\frac {2}{3}}+(sinx)^{\frac {2}{3}}+(sinx)^{\frac {2}{3}} }$

= $\frac {cos x}{(3sinx)^{\frac {2}{3}}}$

= $\frac {cosx(sinx)^{\frac {1}{3}}}{3sinx}$ is the Answer for Given Differentiation!