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agentx5Best ResponseYou've already chosen the best response.2
a la Chain Rule! :D Do you know it?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
If you want you can always replace cos^2(x) with 1sin^2(x), if that's easier for you to derive.
 one year ago

unkabogableBest ResponseYou've already chosen the best response.1
i didn't know that..
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
then you have 12sin^2(x) for your new problem
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
it's still what @agentx5 suggested though,just an algebraic manipulation.
 one year ago

nitzBest ResponseYou've already chosen the best response.1
2cosx(sinx)2sinx(cosx) 2cosxsinx2sinxcosx 4sinxcosx
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
Tada! \(\large \frac{d}{dx}(\cos^2(x)\sin^2(x))\) \(\large \frac{d}{dx}(\cos^2(x))\frac{d}{dx}(\sin^2(x))\) \(\large 2 cos(x) (\frac{d}{dx}(cos(x)))2 sin(x) (\frac{d}{dx}(sin(x)))\) \(\large 2 \cos(x) \sin(x) 2 \sin(x) \cos(x)\) \(\large 4 \sin(x) \cos(x)\)
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
See how the \(\frac{d}{dx}\) is carried from the outer function into the inner? \(\large \frac{d}{dx} (x^3)^2 = 2 (x^3)^{21} \cdot \frac{d}{dx} x^3 = 2 (x^3)^{1} \cdot 3x^{31} \frac{d}{dx} x = 6x^5 \ \ \ \ dx\)
 one year ago

unkabogableBest ResponseYou've already chosen the best response.1
and...voila!! thanks! I'm a fan!!
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
It's the same exact thing, just with sine & cosine instead of exponents or whatever. Nested functions within functions, chain rule. Make sense?
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
In my mind, I treat the derivative as an operand, like a special form of multiplication/division or whathaveyou
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
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: \[(n*u^{n1})*\frac{d}{du}u\] Do the power rule, then take the derivative of what's inside the "shell"/"layer"/"whatever"
 one year ago

agentx5Best ResponseYou've already chosen the best response.2
Learn it well @unkabogable ! It's kind of critical in all Calculus levels (and it only get more complex).
 one year ago
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