Question

Find dy/dx if y= Xpower3+3.x.y+ Y power 3 [derivate]

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {x}^{3} + 3xy + {y}^{3}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( {x}^{3} + 3xy + {y}^{3})$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{3} + \dfrac{d}{dx}3xy + \dfrac{d}{dx}{y}^{3}$

We know,

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

and

$\red{ \boxed{ \sf{ \:\dfrac{d}{dx}uv = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3x}^{2} + 3\bigg(x\dfrac{d}{dx}y + y\dfrac{d}{dx}x \bigg) + {3y}^{2}\dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3x}^{2} + 3\bigg(x\dfrac{dy}{dx} + y \bigg) + {3y}^{2}\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3x}^{2} + 3x\dfrac{dy}{dx} + 3y + {3y}^{2}\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} - 3x\dfrac{dy}{dx} - {3y}^{2}\dfrac{dy}{dx} = {3x}^{2} + 3y$

$\rm :\longmapsto\:\dfrac{dy}{dx}\bigg[1 - {3y}^{2} - 3x\bigg] = 3y + {3x}^{2}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{3y + {3x}^{2} }{1 - 3x - {3y}^{2} }$

Additional Information :-

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