# What is the implicit derivative of 1=xsiny?

Dec 7, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan \frac{y}{x}$

#### Explanation:

If you assume $y$ is a function of $x$ in the equation $1 = x \sin \left(y\right)$, then you need the Product Rule and Chain Rule to differentiate the right-hand side with respect to $x$ to get:

$0 = \frac{d}{\mathrm{dx}} \left(x\right) \cdot \sin \left(y\right) + x \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(y\right)\right) = 1 \cdot \sin \left(y\right) + x \cdot \cos \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

This can then be solved for $\frac{\mathrm{dy}}{\mathrm{dx}}$ by subtracting $\sin \left(y\right)$ from both sides and then dividing both sides by $x \cos \left(y\right)$ to get:

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

A way we can check this answer is to actually solve the original equation for $y$ as a function of $x$:

$1 = x \sin \left(y\right) R i g h t a r r o w \sin \left(y\right) = \frac{1}{x} R i g h t a r r o w y = {\sin}^{- 1} \left(\frac{1}{x}\right) = \arcsin \left(\frac{1}{x}\right)$'

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot - {x}^{- 2} = - \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$

Since $\tan \left(y\right) = \tan \left({\sin}^{- 1} \left(\frac{1}{x}\right)\right) = \frac{1}{\sqrt{{x}^{2} - 1}}$ (draw an appropriate right triangle to help you see this), it follows that

$- \tan \frac{y}{x} = - \frac{1}{x \sqrt{{x}^{2} - 1}} = - \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$