Question

3/(x^2-5)^2 derivative​

Answer 1

Given :

$\\ \red \bigstar \rm \: \frac{3}{ {( {x}^{2} - 5) }^{2} }$

To Find :

• The derivative of given expression =?

Formula :

• Quotient Rule :

Consider a function y=v/x ,in which x >0

By using quotient formula , we get :

$\red \bigstar \boxed{ \sf{ \frac{dy}{dx} = \frac{x \times \frac{d}{dx} v - \frac{d}{dx}x \times v }{ {x}^{2} } }}$

Solution :

• Consider a function y=f(x) ,

$\\ \bullet \sf \: y = \frac{3}{ ({ {x}^{2} - 5)}^{2} }$

Using Quotient rule :

Differentiating both sides with respect to x

$\\ \implies \sf \: y = \frac{3}{ {( {x}^{2} - 5)}^{2} } \\ \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{3}{ {x}^{4} + 25 - 10 {x}^{2} } \\ \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{ {x}^{4} + 25 - 10 {x}^{2} \bigg( \frac{d}{dx} 3 \bigg) - 3 \frac{d}{dx} \bigg( {x}^{4} + 25 - 10 {x}^{2} \bigg)}{( { {x}^{4} + 25 - 10 {x}^{2} )}^{2} } \\ \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{0 - 3(4 {x}^{3} + 0 - 20x) }{ {( {x}^{4} + 25 - 10 {x}^{2} )}^{2} } \\ \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{ - 12 {x}^{3} + 60x}{ { ({x}^{4} + 25 - 10 {x}^{2} )}^{2} }$

The derivative of the given expression is :

$\implies \boxed{ \sf{ \frac{dy}{dx} = \frac{ - 12 {x}^{3} + 60x }{ ({ {x}^{4} + 25 - 10 {x}^{2} )}^{2} } }}$

Additional information :

• Consider a function y=f(x) , y=uv then for solving this type of expression we use product rule :

Product Rule:

$\red\bigstar\boxed{\sf{\dfrac{dy}{dx}= u \dfrac{d}{dv}v+ v \dfrac{d}{du}u}}$

• Consider a function y=v/x then for solving this type of expression we use quotient rule:

Quotient Rule:

$\red\bigstar\boxed{\sf{\dfrac{dy}{dx}=\dfrac{x \dfrac{d}{dx}v- v \dfrac{d}{dx}}{x^{2}}}}$