Question

x^2+6xy+y^2=10
To show d^2y/dx^2=80/(3x+y)^3

Answer 1

${\bold{\huge{\textbf{\red{Answer}}}}}$

Given Ques:

${\large{\bold{x^2+6xy+y^2=10}}}\rightarrow(1)$

To show

${\bold{\red{\frac{d^2y}{dx^2}=\frac{80}{(3x+y)}^3}}}$

differentiating equation (1) with respect to x

${2x+ [6y+6x \frac{dy}{dx}]+ 2y \frac{dy}{dx}=0}$

or

${\frac{2x+6y}{-6x-2y}= \frac{dy}{dx}}$

taking common:

${\frac{x+3y}{-3x-y}= \frac{dy}{dx}}$

by calculas divident rule:

$</strong><strong>{</strong><strong>{\bold{\red{\frac{d^2y}{dx^2}=\frac{80}{(3x+y)}^3}}</strong><strong>}</strong><strong>}$