Question

differentiate:
y=(x²+1)(x/2+1)​

Answer 1

Solution :

$\sf{\dfrac{dy}{dx} = \dfrac{x^{2} + 1}{\dfrac{x}{2} + 1}}$

By using the quotient rule of differentiation, we get :

• Quotient rule of differentiation :

$\sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\cdot\dfrac{d(u)}{dx} - (u)\cdot\dfrac{d(v)}{dx}}{v^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{x}{2} + 1\bigg)\cdot\dfrac{d(x^{2} + 1)}{dx} - (x^{2} + 1)\cdot\dfrac{d\bigg(\dfrac{x}{2} + 1\bigg)}{dx}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{x}{2} + 1\bigg)\cdot\dfrac{d(x^{2})}{dx} + \dfrac{d(1)}{dx} - (x^{2} + 1)\cdot\dfrac{1}{2}\cdot\dfrac{d(x)}{dx} + \dfrac{d(1)}{dx}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{x}{2} + 1\bigg)\cdot (2x) + 0 - (x^{2} + 1)\cdot\dfrac{1}{2}\cdot(1) + 0}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{x}{2} + 1\bigg)\cdot (2x) - (x^{2} + 1)\cdot\dfrac{1}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{2x^{2}}{2} + 2x\bigg) - (x^{2} + 1)\cdot\dfrac{1}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\bigg(\dfrac{2x^{2}}{2} + 2x\bigg) - \dfrac{(x^{2} + 1)}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\dfrac{(2x^{2} + 4x) - (x^{2} + 1)}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\dfrac{2x^{2} + 4x - x^{2} - 1}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{\dfrac{x^{2} + 4x - 1}{2}}{\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{x^{2} + 4x - 1}{2\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}$

$\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{x^{2} + 4x - 1}{2\bigg(\dfrac{x}{2} + 1\bigg)^{2}}}}$