# Help with implicit differentiation: $e^{9x}= \sin(x+9y)$

Find $\;\dfrac{dy}{dx}\;$ given $\;e^{9x}= \sin(x+9y)$

the answer is $\;\displaystyle\frac{e^{9x}}{\cos(x+9y)}- \frac{1}{9}$.

Can you show the process of how this is worked?

thanks.

Differentiate both sides of your equation using chain rule $$\left( e^{9x} \right )' = \left ( \sin (x+9y) \right )'\\ 9e^{9x} = \cos (x+9y) \cdot (x + 9y)' = \cos (x+9y) \cdot (1+9y')$$ From which you can find that $$1+9y' = \frac {9e^{9x}}{\cos (x+9y)} \\ y' = \frac {e^{9x}}{\cos (x+9y)}-\frac 19$$
Let's start by taking derivatives with respect to $x$ at both sides of the equation. We get
$$\frac{d}{dx}(e^{9x}) = \frac{d}{dx}(\sin(x + 9y))$$ which gives, by applying the chain rule and remembering that $y$ is a function dependent on $x$ $$9e^{9x} = \cos(x + 9y)(1 + 9\frac{dy}{dx})$$ Now dividing both sides by $\cos(x + 9y)$ gives $$\frac{9e^{9x}}{\cos(1+9y)}= 1 + 9\frac{dy}{dx}$$ from which we can deduce $$\frac{9e^{9x}}{\cos(1+9y)} - 1 = 9\frac{dy}{dx}$$ and so $$\frac{e^{9x}}{\cos(1+9y)} - \frac{1}{9} = \frac{dy}{dx}$$
Taking the derivative with respect to $x$ on the left-hand side is basic stuff. For the right-hand side, use the chain rule (with the assumption that $y$ is a function of $x$) to get \begin{align}\frac{d}{dx}\bigl[\sin(x+9y)\bigr] &= \cos(x+9y)\cdot\frac{d}{dx}[x+9y]\\ &= \cos(x+9y)\cdot\left(1+9\frac{dy}{dx}\right).\end{align} Now set your left-hand side's derivative equal to your right-hand side's derivative, and solve for $\cfrac{dy}{dx}.$