Question

$\frac{d}{dx} = ( {x}^{3} + x + 1)^ \frac{4}{3}$
Find the derivative ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{y=\left({x}^{3}+x+1\right)^{\frac{4}{3}}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{\left({x}^{3}+x+1\right)^{\frac{4}{3}}\right\}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{4}{3}\left({x}^{3}+x+1\right)^{\frac{4}{3}-1}\cdot\dfrac{d}{dx}\left({x}^{3}+x+1\right)}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{4}{3}\left({x}^{3}+x+1\right)^{\frac{1}{3}}\cdot\left(3{x}^{2}+1\right)}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{12{x}^{2}+4}{3}\cdot\sqrt[3]{{x}^{3}+x+1}}$