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

2 Answers
Nov 4, 2017

#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 :)

Nov 4, 2017

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!