What is the implicit derivative of 1=sin(2x)/y?

Feb 26, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y \cos \left(2 x\right)}{\sin} \left(2 x\right) = 2 y \cot \left(2 x\right)$

Explanation:

Differentiate both sides with respect to $x$ and solve for $\frac{\mathrm{dy}}{\mathrm{dx}} .$ Differentiating with respect to $x$ means that every time we differentiate a term containing $y$, we must end up with an instance of $\frac{\mathrm{dy}}{\mathrm{dx}} :$

$\frac{d}{\mathrm{dx}} 1 = \frac{d}{\mathrm{dx}} \left(\sin \frac{2 x}{y}\right)$

$\frac{d}{\mathrm{dx}} \left(1\right) = 0$, so the left side becomes $0.$

Differentiate the right side using the quotient rule, $\left(\frac{f}{g}\right) ' = \frac{g f ' - f g '}{g} ^ 2$

$0 = \frac{y \left(2 \cos \left(2 x\right)\right) - \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(2 x\right)}{y} ^ 2$

We want to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$. Multiply both sides by ${y}^{2}$. $0 \cdot {y}^{2} = 0$, so the left side is still $0.$

$0 = 2 y \cos \left(2 x\right) - \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(2 x\right) = 2 y \cos \left(2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y \cos \left(2 x\right)}{\sin} \left(2 x\right) = 2 y \cot \left(2 x\right)$ (as $\cos \frac{x}{\sin} \left(x\right) = \cot \left(x\right)$

We'll keep this implicit solution containing $y$ as we want the implicit derivative.