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

##### 2 Answers

#### 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

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
#dy/dx# at the end. - Solve for
#dy/dx# .

So, let's dive into this problem:

**Step #1: Take Derivatives of Both Sides of the Equation**

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

**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:

*note that

**Step 4: Solve for #dy/dx#:**

And there's your final answer!

Here's some videos that might help:

Hope that helped :)

Alternatively:

#cosy = e^(2x+ 5)#

We know that

#-siny(dy/dx) = 2e^(2x + 5)#

#dy/dx = (2e^(2x + 5))/(-siny)#

Since

#dy/dx= (2e^(2x+ 5))/-sqrt(1 - cos^2y)#

#dy/dx= -(2e^(2x + 5))/sqrt(1 - (e^(2x +5))^2)#

Hopefully this helps!