Question

differenciate
$y = cosx \div {x}^{3}$

Answer 1

Hey there!

Solution:


$Let \:y = \dfrac{cosx}{x^{3}}\\ \\ Now \: Differentiate\: Both \:sides\: wrt \:x\\ \\ \dfrac{dy}{dx} = \dfrac{x^{3} ( -sinx) - 3x^{2} (cosx)}{x^{6}} \\ \\ \dfrac{dy}{dx} = \dfrac{-x^{3} sinx - 3x^{2} cosx}{x^{6}} \\ \\ Taking\: -\: common\: from\: the \:R.H.S.\\ \\ \dfrac{dy}{dx} = - \dfrac{(x^{3}sinx + 3x^{2}cosx)}{x^{6}}\\ \\ Now\: taking\: x^{2} \:common\: from \:R.H.S \:again\\ \\ \dfrac{dy}{dx} =- \dfrac{x^{2} ( xsinx + 3cosx)}{x^{6}} \\ \\ \dfrac{dy}{dx} = - \dfrac{xsinx + 3cosx}{x^{4}} \\ \\ Points\: to \:Remember\::\\ \\ \dfrac{d(cosx)}{dx} = -sinx$


This sum is solved by $\textbf{Quotient rule}$.


#Be Brainly.

Answer 2

$\tt \: y = \frac{cosx}{ {x}^{3} }$

Differentiate both the sides.

$\tt \: \frac{dy}{dx} = \frac{ { - x}^{3} sinx - {3x}^{2} cosx}{ {x}^{6}}$

For the rest refer the attachment.

#Akshi❣️