Question

1. y =12x^5 +3x^4+ 7x³+x² -9x
differesitation​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Differentiate with respect to x,

$\rm :\longmapsto\:y = {12x}^{5} + {3x}^{4} + {7x}^{3} + {x}^{2} - 9x$

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

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

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

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

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

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

Given that

$\rm :\longmapsto\:y = {12x}^{5} + {3x}^{4} + {7x}^{3} + {x}^{2} - 9x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( {12x}^{5} + {3x}^{4} + {7x}^{3} + {x}^{2} - 9x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{12x}^{5}+\dfrac{d}{dx}{3x}^{4}+\dfrac{d}{dx}{7x}^{3}+\dfrac{d}{dx}{x}^{2}-\dfrac{d}{dx}9x$

$\rm :\longmapsto\:\dfrac{dy}{dx} =12\dfrac{d}{dx}{x}^{5}+3\dfrac{d}{dx}{x}^{4}+7\dfrac{d}{dx}{x}^{3}+\dfrac{d}{dx}{x}^{2}-9\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {12(5)x}^{5 - 1} + 3(4) {x}^{4 - 1} + 7(3) {x}^{3 - 1} + {2x}^{2 - 1} - 9$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {60x}^{4} +12{x}^{3} +21{x}^{2}+{2x}-9$

Additional Information :-

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

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

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

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

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

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