How do you find dy/dx by implicit differentiation for 1+x = sin(xy^2)?

May 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos \left(x \cdot {y}^{2}\right) \cdot 2 x \cdot y} - \frac{y}{2 x}$

Explanation:

Implicitly differentiate the Left-Hand Side with respect to $x$:

• $\frac{d}{\mathrm{dx}} \left[1 + x\right] = 1$ by the power rule

The right-hand side is slightly trickier...

• $\textcolor{w h i t e}{l} \frac{d}{\mathrm{dx}} \left[\sin \left(x \cdot {y}^{2}\right)\right]$
$= \cos \left(x \cdot {y}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left[x \cdot {y}^{2}\right]$
$= \cos \left(x \cdot {y}^{2}\right) \cdot \left({y}^{2} + 2 x \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
by the chain rule along with the cosine rule.

Hence

$\frac{d}{\mathrm{dx}} \left[1 + x\right] = \frac{d}{\mathrm{dx}} \left[\sin \left({x}^{2} \cdot y\right)\right]$
$1 = \cos \left(x \cdot {y}^{2}\right) \cdot \left({y}^{2} + 2 x \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

The question is asking for $\textcolor{\mathrm{da} r k b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}}$ so we shall seek to isolate that expression as a whole.

$1 = \cos \left(x \cdot {y}^{2}\right) \cdot \left({y}^{2}\right) + \cos \left(x \cdot {y}^{2}\right) \cdot 2 x \cdot y \cdot \textcolor{\mathrm{da} r k b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}}$
$\cos \left(x \cdot {y}^{2}\right) \cdot 2 x \cdot y \cdot \textcolor{\mathrm{da} r k b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = 1 - \cos \left(x \cdot {y}^{2}\right) \cdot \left({y}^{2}\right)$

$\textcolor{\mathrm{da} r k b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{\cos \left(x \cdot {y}^{2}\right) \cdot 2 x \cdot y} - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos \left(x \cdot {y}^{2}\right)}}} \cdot {y}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos \left(x \cdot {y}^{2}\right)}}} \cdot 2 x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{y}}}}$
$\textcolor{w h i t e}{\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}}} = \frac{1}{\cos \left(x \cdot {y}^{2}\right) \cdot 2 x \cdot y} - \frac{y}{2 x}$