Question

If $x^3{y}^4=(x+y{)}^7$ then find $\frac{dy}{dx}$.

Answer 1

Given,

$x^3{y}^4=(x+y{)}^7$

Taking logarithm both sides with respect to the base $e$ we get,

$3\log x+4\log y=7\log (x+y)$

Now differentiating both sides with respect to $x$ we get,

$\frac{3}{x}+\frac{4}{y}\frac{dy}{dx}=\frac{7}{(x+y)}(1+\frac{dy}{dx})$

or, $(\frac{3}{x}-\frac{7}{x+y})$$=(\frac{7}{x+y}-\frac{4}{y})\frac{dy}{dx}$

or, $\frac{dy}{dx}=\frac{y(3y-4x)}{x(3y-4x)}=\frac{y}{x}$