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

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agentx5 Group TitleBest ResponseYou've already chosen the best response.2
a la Chain Rule! :D Do you know it?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

unkabogable Group TitleBest ResponseYou've already chosen the best response.1
i didn't know that..
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
then you have 12sin^2(x) for your new problem
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
it's still what @agentx5 suggested though,just an algebraic manipulation.
 2 years ago

nitz Group TitleBest ResponseYou've already chosen the best response.1
2cosx(sinx)2sinx(cosx) 2cosxsinx2sinxcosx 4sinxcosx
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.2
Tada! \(\large \frac{d}{dx}(\cos^2(x)\sin^2(x))\) \(\large \frac{d}{dx}(\cos^2(x))\frac{d}{dx}(\sin^2(x))\) \(\large 2 cos(x) (\frac{d}{dx}(cos(x)))2 sin(x) (\frac{d}{dx}(sin(x)))\) \(\large 2 \cos(x) \sin(x) 2 \sin(x) \cos(x)\) \(\large 4 \sin(x) \cos(x)\)
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.2
See how the \(\frac{d}{dx}\) is carried from the outer function into the inner? \(\large \frac{d}{dx} (x^3)^2 = 2 (x^3)^{21} \cdot \frac{d}{dx} x^3 = 2 (x^3)^{1} \cdot 3x^{31} \frac{d}{dx} x = 6x^5 \ \ \ \ dx\)
 2 years ago

unkabogable Group TitleBest ResponseYou've already chosen the best response.1
and...voila!! thanks! I'm a fan!!
 2 years ago

agentx5 Group TitleBest 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?
 2 years ago

agentx5 Group TitleBest 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
 2 years ago

agentx5 Group TitleBest 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"
 2 years ago

agentx5 Group TitleBest 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).
 2 years ago
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