Question

Differentiate ∛ x³ + 3x² -7x +1 using the chain rule.

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

$\boxed{\bf\:\dfrac{d}{dx}k = 0}$

$\boxed{\bf\:\dfrac{d}{dx}x= 1}$

$\boxed{\bf\:\dfrac{d}{dx}k \: f(x)= k \: \dfrac{d}{dx}f(x)}$

Solution :-

$\rm :\longmapsto\:Let \: y = \sqrt[3]{ {x}^{3} + {3x}^{2} - 7x + 1 }$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg( {x}^{3} + {3x}^{2} - 7x + 1\bigg) }^{\dfrac{1}{3} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}{\bigg( {x}^{3} + {3x}^{2} - 7x + 1\bigg) }^{\dfrac{1}{3}}$

$\rm =\dfrac{1}{3}{\bigg( {x}^{3}+{3x}^{2}-7x+1\bigg) }^{\dfrac{1}{3} - 1}\dfrac{d}{dx}( {x}^{3} +{3x}^{2}-7x + 1)$

$\rm =\dfrac{1}{3}{\bigg( {x}^{3}+{3x}^{2}-7x+1\bigg) }^{\dfrac{1-3}{3}}\bigg(\dfrac{d}{dx}{x}^{3}+3\dfrac{d}{dx}{x}^{2} -7\dfrac{d}{dx}x+\dfrac{d}{dx}1\bigg)$

$\rm =\dfrac{1}{3}{\bigg( {x}^{3}+{3x}^{2}-7x+1\bigg) }^{ - \dfrac{2}{3}}(3 {x}^{2}+3\times 2x-7+0)$

$\rm =\dfrac{1}{3}{\bigg( {x}^{3}+{3x}^{2}-7x+1\bigg) }^{ - \dfrac{2}{3}}(3 {x}^{2}+6x-7)$

Additional Information :-

$\boxed{\bf\:\dfrac{d}{dx}sinx = cosx}$

$\boxed{\bf\:\dfrac{d}{dx}cosx = - sinx}$

$\boxed{\bf\:\dfrac{d}{dx}cosecx = - cosecx \: cotx}$

$\boxed{\bf\:\dfrac{d}{dx}secx = secx \: tanx}$

$\boxed{\bf\:\dfrac{d}{dx}tanx = {sec}^{2}x}$

$\boxed{\bf\:\dfrac{d}{dx}cotx = { - \: cosec}^{2}x}$

$\boxed{\bf\:\dfrac{d}{dx} {e}^{x} = {e}^{x}}$

$\boxed{\bf\:\dfrac{d}{dx} {a}^{x} = {a}^{x}loga}$

$\boxed{\bf\:\dfrac{d}{dx}logx = \dfrac{1}{x}}$