Question

Differentiate:
2x^3+5xy+6y^2=87

Answer 1

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

Given function is

$\rm \: {2x}^{3} + 5xy + {6y}^{2} = 87$

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

$\rm \: \dfrac{d}{dx}\bigg[{2x}^{3} + 5xy + {6y}^{2}\bigg] = \dfrac{d}{dx}87$

We know,

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

So, using this, we get

$\rm \: 2\dfrac{d}{dx} {x}^{3} + 5\dfrac{d}{dx}xy + 6\dfrac{d}{dx} {y}^{2} = 0$

We know,

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

and

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

So, using this, we get

$\rm \: 2 \times {3x}^{3 - 1} + 5\bigg(x\dfrac{d}{dx}y + y\dfrac{d}{dx}x\bigg) + 6 \times {2y}^{2 - 1}\dfrac{dy}{dx} = 0$

$\rm \: {6x}^{2} + 5\bigg(x\dfrac{dy}{dx} + y \times 1\bigg) + 12y\dfrac{dy}{dx} = 0$

$\rm \: {6x}^{2} + 5x\dfrac{dy}{dx} + 5y + 12y\dfrac{dy}{dx} = 0$

$\rm \: (5x +12y)\dfrac{dy}{dx} = - ( {6x}^{2} + 5y)$

$\rm\implies \: \: \boxed{\tt{ \: \dfrac{dy}{dx} = - \dfrac{( {6x}^{2} + 5y)}{(5x +12y)} \: \: }}$

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ADDITIONAL INFORMATION

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