Question

(b) If x2 + 2xy + y3 = 42, find dy/dx​

Answer 1

$\huge\boxed{ \green{\mathfrak{solution}}}$

Given:

$\bold{ {x}^{2} + 2xy + {y}^{3} = 42}$

Differentiate w.r.t x both sides,

$\frac{d}{dx} ( {x}^{2} + 2xy + {y}^{3} ) = \frac{d}{dx} 42 \\ \\ \rightarrow \: \frac{d}{dx} {x}^{2} + \frac{d}{dx} 2xy + \frac{d}{dx} {y}^{3} = 0 \\ \\ \rightarrow \: 2x + 2(x \: \frac{dy}{dx} + y) + 3 {y}^{2} \frac{dy}{dx} = 0 \\ \\ \rightarrow \: 2x + 2y + \frac{dy}{dx} (2x + 3 {y}^{2} ) = 0 \\ \\ \rightarrow \: \frac{dy}{dx} (2x + 3 {y}^{2} ) = - 2(x + y) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{ - 2(x + y)}{2x + 3 {y}^{2} }$

Formula :

◆ d /dx(x^n)= nx^(n-1)

◆ d/dx(u.v) = u dv/dx+ v du/dx