Question

apply Quotient Rule and find dy/dx​

Answer 1

Question:-

Apply Quotient rule to find dy/dx

$\rm \: y = \dfrac{ {x}^{3} - 7}{2 - {x}^{2} }$

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

Given function is

$\rm \: y = \dfrac{ {x}^{3} - 7}{2 - {x}^{2} }$

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

$\rm \: \dfrac{d}{dx}y =\dfrac{d}{dx} \: \dfrac{ {x}^{3} - 7}{2 - {x}^{2} }$

We know,

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} \frac{u}{v} \: = \: \dfrac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: \: }}$

So, here

$\rm \: u = {x}^{3} - 7$

and

$\rm \: v = 2 - {x}^{2}$

So, on substituting the values, we get

$\rm \: \dfrac{dy}{dx}$

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

$\rm \: =  \: \dfrac{(2 - {x}^{2})(3 {x}^{2} - 0) - ( {x}^{3} - 7)(0 - 2x)}{ {(2 - {x}^{2} )}^{2} }$

$\rm \: =  \: \dfrac{(2 - {x}^{2})(3 {x}^{2}) - ( {x}^{3} - 7)(- 2x)}{ {(2 - {x}^{2} )}^{2} }$

$\rm \: =  \: \dfrac{ {6x}^{2} - 3{x}^{4} + 2{x}^{4} - 14x}{ {(2 - {x}^{2} )}^{2} }$

$\rm \: =  \: \dfrac{ { - x}^{4} + 6{x}^{2} - 14x}{ {(2 - {x}^{2} )}^{2} }$

$\rm \: =  \: \dfrac{ -x ( {x}^{3} - 6{x} + 14)}{ {(2 - {x}^{2} )}^{2} }$

Hence,

$\rm\implies \: \boxed{\sf{  \:\: \rm \:\dfrac{dy}{dx} =  \: \dfrac{ -x ( {x}^{3} - 6{x} + 14)}{ {(2 - {x}^{2} )}^{2} } \: \: }}$

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Formulae Used :-

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

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

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$