Question

If $x^{16}{y}^9={(x^2+y)}^{17}$, prove that $\frac{dy}{dx}=\frac{2y}{x}$.

Answer 1

$x^{16}{y}^{16}=(x^2+y{)}^{17}$

Taking $\log$

$16\log x+9\log y=17\log (x^2+y)$

Differentiating both sides.

$\Rightarrow \frac{16}{x}+\frac{9}{y}.y^{\prime }=\frac{17}{x^2+y}(2x+y^{\prime })$

$\Rightarrow \frac{16}{x}-\frac{34x}{x^2+y}=y^{\prime }(\frac{17}{x^2+y}-\frac{9}{4})$

$\Rightarrow \frac{2(-9x^2+8y)}{x}=y^{\prime }\left(\frac{-9x^2+8y}{y}\right)$

$\Rightarrow y^{\prime }=\frac{\left(2y\right)}{\left(x\right)}$