Question

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

Answer 1

`dy/dx = 2/-tan(y)`

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 `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

`=> d/dxln(cos(y)) = d/dx(2x+5)`

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

`=> d/dxln(cos(y)) = 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:

`=> 1/cos(y) * -sin(y) * dy/dx = 2`

`=> -tan(y)dy/dx = 2`

*note that `sin(x)/cos(x)` evaluates to `tan(x)`.

Step 4: Solve for `dy/dx`:

`=> dy/dx = 2/-tan(x)`

And there's your final answer!

Here's some videos that might help:

Hope that helped :)

Answer 2

Alternatively:

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

We know that `d/dx(e^(f(x))) = f'(x)e^(f(x))`. Thus:

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

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

Since `cosy = sqrt(1 - sin^2y)`

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

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

Hopefully this helps!