# How do you find dy/dx by implicit differentiation given ln(cosy)=2x+5?

Nov 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{-} \tan \left(y\right)$

#### Explanation:

Implicit Differentiation helps you take derivatives of functions that have different variables from the one with respect to which you are taking the derivative (usually, when you have $y$'s in your function).

The steps to this are simple:

• Take any derivatives with $x$'s in them as normal.
• When you have a $y$:
• Take the derivaive as normal BUT:
• Tag on a $\frac{\mathrm{dy}}{\mathrm{dx}}$ at the end.
• Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

So, let's dive into this problem:

Step 1: Take Derivatives of Both Sides of the Equation

$\implies \frac{d}{\mathrm{dx}} \ln \left(\cos \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(2 x + 5\right)$

Step 2: Evaluate Derivatives of any "x" terms:

$\implies \frac{d}{\mathrm{dx}} \ln \left(\cos \left(y\right)\right) = 2$

Step #3: Evaluate Derivatives of any "y" terms:

You'll need to use a chain rule to evaluate this, but it's a very simple one:

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

$\implies - \tan \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

*note that $\sin \frac{x}{\cos} \left(x\right)$ evaluates to $\tan \left(x\right)$.

Step 4: Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{-} \tan \left(x\right)$

Here's some videos that might help:

Hope that helped :)

Nov 4, 2017

Alternatively:

$\cos y = {e}^{2 x + 5}$

We know that $\frac{d}{\mathrm{dx}} \left({e}^{f \left(x\right)}\right) = f ' \left(x\right) {e}^{f \left(x\right)}$. Thus:

$- \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 {e}^{2 x + 5}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{2 x + 5}}{- \sin y}$

Since $\cos y = \sqrt{1 - {\sin}^{2} y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{2 x + 5}}{-} \sqrt{1 - {\cos}^{2} y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 {e}^{2 x + 5}}{\sqrt{1 - {\left({e}^{2 x + 5}\right)}^{2}}}$

Hopefully this helps!