Question

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

Answer 1

a la Chain Rule! :-D

Do you know it?

Answer 2

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

Answer 3

i didn't know that..

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

See how the \(\frac{d}{dx}\) is carried from the outer function into the inner?

\(\large \frac{d}{dx} (x^3)^2 = 2 (x^3)^{2-1} \cdot \frac{d}{dx} x^3 = 2 (x^3)^{1} \cdot 3x^{3-1} \frac{d}{dx} x = 6x^5 \ \ \ \ dx\)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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"

Answer 13

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