## unkabogable 2 years ago how to get for the derivative of cos^2 (x)- sin^2 (x)?

1. agentx5

a la Chain Rule! :-D Do you know it?

2. Spacelimbus

If you want you can always replace cos^2(x) with 1-sin^2(x), if that's easier for you to derive.

3. unkabogable

i didn't know that..

4. Spacelimbus

then you have 1-2sin^2(x) for your new problem

5. Spacelimbus

it's still what @agentx5 suggested though,just an algebraic manipulation.

6. nitz

2cosx(-sinx)-2sinx(cosx) -2cosxsinx-2sinxcosx -4sinxcosx

7. agentx5

Tada! $$\large \frac{d}{dx}(\cos^2(x)-\sin^2(x))$$ $$\large \frac{d}{dx}(\cos^2(x))-\frac{d}{dx}(\sin^2(x))$$ $$\large 2 cos(x) (\frac{d}{dx}(cos(x)))-2 sin(x) (\frac{d}{dx}(sin(x)))$$ $$\large -2 \cos(x) \sin(x) -2 \sin(x) \cos(x)$$ $$\large -4 \sin(x) \cos(x)$$

8. agentx5

See how the $$\frac{d}{dx}$$ is carried from the outer function into the inner? $$\large \frac{d}{dx} (x^3)^2 = 2 (x^3)^{2-1} \cdot \frac{d}{dx} x^3 = 2 (x^3)^{1} \cdot 3x^{3-1} \frac{d}{dx} x = 6x^5 \ \ \ \ dx$$

9. unkabogable

and...voila!! thanks! I'm a fan!!

10. agentx5

It's the same exact thing, just with sine & cosine instead of exponents or whatever. Nested functions within functions, chain rule. Make sense?

11. agentx5

In my mind, I treat the derivative as an operand, like a special form of multiplication/division or what-have-you

12. agentx5

Granted that's not TECHNICALLY what it is, but you can "distribute" across added/subtracted terms and it applies into nested functions with the chain rule... Chain Rule: for: $u^n$ then: $(n*u^{n-1})*\frac{d}{du}u$ Do the power rule, then take the derivative of what's inside the "shell"/"layer"/"whatever"

13. agentx5

Learn it well @unkabogable ! It's kind of critical in all Calculus levels (and it only get more complex).