calculate the derivative and simplify h(x)=3xsin^2 (x) + cos^3 (x)
Let's start with the first term.
How do differentiate 3xsin^2 (x)?
product rule?
Correct
d/dx(3x)sin(x)+3xd/dx(sinx)
good, so what do you get if simplified?
3sin^2(x)+3cos^2(x)
wait. Sorry. You're not quite there.
\[\frac{ d }{ dx }3xsin^2x=(\frac{ d }{ dx }3x)\sin^2x+3x(\frac{ d }{ dx }\sin^2x)\]
\[=3\sin^2x+3x(2\sin x \cos x)=3\sin^2x+6x \sin x \cos x\]
Do you know the formula for differentiating composite function?
is that the chain rule?
Correct. I think you forgot to use it when differentiating sin^2x ;)
Yeah i get confused when it needs to be used. Does it always have to used with trig functions?
no. You always use it when differentiating powers of a function. So for y={f(x)}^n, chain rule is always used.
Ok yeah that makes sense.
Now, using that, try differentiating the second term of the original problem.
So would d/dx cos^3(x) be 3cos(x)*-sin(x)
Close. y={f(x)}^n y'=n{f(x)}^(n-1)*f'(x)
3cos^2(x)-sin(x)
correct. Now add those 2 terms.
-3cos^2(x)sin(x) is that what you are meaning?
Remember how we differentiated the first term? Add that and the one you just found.
k i wasn't sure which one you meant 3sin^2(x)+3x(sin(x)cos(x)-3cos^2(x)sin(x)
What was the answer for the differentiation of the first term?
wasn't it 3sin^2(x)+3x(2sin(x)cos(x))
Oh. I see. In my personal opinion, I think you should write it as 3sin^2(x)+6xsin(x)cos(x), but it's your choice.
Now that you got it, add it with your 2nd term. What do you get?
3sin^2(x)+6xsin(x)cos(x)-3cos^2(x)sin(x) would the next step be to factor out a 3?
Actually, you can just leave it like that. Say hi to your answer :)
It says i have to simplify?
is that as simplified as it can get?
Yep!
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