# What is the derivative of tan(x+y) = ln x + 5y?

Dec 13, 2017

dy/dx = (xsec^2(x+y) - 1)/(5x - xsec^2(x+y)

#### Explanation:

Since you have an equation mixed with $y$'s and $x$'s, you know that the key concept at play here is implicit differentiation. If you're not sure what that really is/does, I'd recommend watching my video on it:

Anyways, the 3 key steps to implicit differentiation are:

• Differentiate terms with an $x$ in them as usual.
• Differentiate terms with a $y$ in them as normal.
• BUT tag on a $\frac{\mathrm{dy}}{\mathrm{dx}}$ to whatever you get.
• Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

So, with this in mind, we start by taking the derivative of both sides of this equation (with respect to $x$):

$\frac{d}{\mathrm{dx}} \left(\tan \left(x + y\right)\right) = \frac{d}{\mathrm{dx}} \left(\ln x\right) + \frac{d}{\mathrm{dx}} \left(5 y\right)$

I'm not going to walk through the intricacies of actually doing the derivatives, since this is not a focus of this problem, but know that you will need a chain rule to evaluate the derivative of the left hand side of that equation.

You should get:

${\sec}^{2} \left(x + y\right) \cdot \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{x} + 5 \frac{\mathrm{dy}}{\mathrm{dx}}$

Now, I'm going to distribute my ${\sec}^{2} \left(x\right)$ to both the $1$ and the $\frac{\mathrm{dy}}{\mathrm{dx}}$:

${\sec}^{2} \left(x + y\right) + {\sec}^{2} \left(x + y\right) \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{x} + 5 \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}}}$

The reason I did this is just so that I have two single terms with a $\frac{\mathrm{dy}}{\mathrm{dx}}$ in them that I can work with. Now, I will move all my $\frac{\mathrm{dy}}{\mathrm{dx}}$ term to one side, and my other terms to the other:

${\sec}^{2} \left(x + y\right) - \frac{1}{x} = 5 \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}}} - {\sec}^{2} \left(x + y\right) \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}}}$

Now, we factor out the $\frac{\mathrm{dy}}{\mathrm{dx}}$, and solve:

color(red)(dy/dx) = (sec^2(x+y) - 1/x)/(5 - sec^2(x+y)

...and that is a perfectly good final answer. However, fractions in fractions aren't nice to look at, so we're going to multiply through by $\frac{x}{x}$ to get rid of that $\frac{1}{x}$:

color(red)(dy/dx) = (xsec^2(x+y) - 1)/(5x - xsec^2(x+y)

This would be the final answer that you'd most likely see in your textbook, or in other literary sources. Note that it's perfectly OK to leave your answer in terms of $x$ and $y$ in implicit differentiation problems.

Hope that helped :)