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mathcalculus
HELP! who can explain derivatives :( let f(x)= (x^3+3x+8)^2 F'(x)= F'(2)=
use the chain rule get \[F'(x)=2(x^2+3x+8)(3x^2+3)\]
I am very bad with derivatives. I missed the whole lesson de to being sick. can we start from the beginning?
how do we use the chain rule?
you have something squared the derivative of something squared is two times something, times the derivative of something
for example, the derivative of \(\sin^2(x)\) is \(2\sin(x)\cos(x)\)
the derivative of \(x+\sqrt{x}+\tan(x))^2\) is \[2(x+\sqrt{x}+\tan(x))(1+\frac{1}{2\sqrt{x}}+\sec^2(x))\]
so how do we begin this problem?
to find the derivative we need to place the exponent in front... correct?
is the derivative for f'(x) = 2X^2+6X+16?