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

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

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

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

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

nitz
 2 years ago
Best ResponseYou've already chosen the best response.12cosx(sinx)2sinx(cosx) 2cosxsinx2sinxcosx 4sinxcosx

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

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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.2It's the same exact thing, just with sine & cosine instead of exponents or whatever. Nested functions within functions, chain rule. Make sense?

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.2In my mind, I treat the derivative as an operand, like a special form of multiplication/division or whathaveyou

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.2Granted 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"

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.2Learn it well @unkabogable ! It's kind of critical in all Calculus levels (and it only get more complex).
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