Implicit Differentiation of a Trig Function?

May 2010
11
1
For some reason, I'm having A LOT of problems differentiating trig functions, whether or not they're implicit.

sin x + 2 cos 2y = 1

I got as far as cos x - 2 sin 2y... and then I feel like there's another term added on here.

I have the answer (from the book), I just don't have the middle steps to understand what happened.
 

TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
For some reason, I'm having A LOT of problems differentiating trig functions, whether or not they're implicit.

sin x + 2 cos 2y = 1

I got as far as cos x - 2 sin 2y... and then I feel like there's another term added on here.

I have the answer (from the book), I just don't have the middle steps to understand what happened.
Taking the implicit derivative gives

\(\displaystyle \cos(x)-4\sin(2y)\frac{dy}{dx}=0\)
 
May 2010
11
1
But where does the 4(sin y) come from?

For example, my book writes this:

sin x + 2 cos 2y = 1
cos x - 4(sin 2y)dy/dx = 0
dy/dx = cos x/4 sin 2y

I understand how to get from step 2 to step 3, and I understand part of step two (cos x is the derivative of sin x, so that makes sense, and the negative part of the next part comes from the fact that the derivative of cos x is -sin x), but how exactly does the 4(sin 2y) come about?
 

TheEmptySet

MHF Hall of Honor
Feb 2008
3,764
2,029
Yuma, AZ, USA
But where does the 4(sin y) come from?

For example, my book writes this:

sin x + 2 cos 2y = 1
cos x - 4(sin 2y)dy/dx = 0
dy/dx = cos x/4 sin 2y

I understand how to get from step 2 to step 3, and I understand part of step two (cos x is the derivative of sin x, so that makes sense, and the negative part of the next part comes from the fact that the derivative of cos x is -sin x), but how exactly does the 4(sin 2y) come about?
You need to use the chain rule don't forget to take the derivative of the inside

\(\displaystyle \frac{d}{dx}f(2y)=\frac{df}{dy}(2\frac{dy}{dx})\)

\(\displaystyle \frac{d}{dx}2\cos(2y)=2(-\sin(2y))\frac{d}{dx}(2y)=-4\sin(2y)\frac{dy}{dx}\)
 
  • Like
Reactions: Ohoneo
May 2010
11
1
Ah, thank you so much. That's what I was missing. Thank you!