Thread: Implicit Differentiation of a Trig Function?

1. Implicit Differentiation of a Trig Function?

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.

2. Originally Posted by Ohoneo
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

$\cos(x)-4\sin(2y)\frac{dy}{dx}=0$

3. 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?

4. Originally Posted by Ohoneo
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

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

$\frac{d}{dx}2\cos(2y)=2(-\sin(2y))\frac{d}{dx}(2y)=-4\sin(2y)\frac{dy}{dx}$

5. Ah, thank you so much. That's what I was missing. Thank you!