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Find the indicated derivative of the function f(x) = cosx^2, find f''(x)

Mathematics
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\[2\cos(x)\sin(x)\]is your first derivative take the derivative of that one, using the product rule
@satellite73 it will be -2cos(x)sin(x)
@genius12 so the derivative of -2cos(x) sin(x) , and i got -2 cos(2x)

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Other answers:

@jazy should i now take the derivative of -2 cos(2x) ? or
which one : |dw:1362427439289:dw|
cosx^2
so, the first one in my diagram ?
yes
oke, to get the first derivative of f(x) we can use the cain rule hint : derivative of x^2 = 2x derivative of cos = - sin
yea i got it the first derivative will be -2 cos(x) sin (x) right?
so, derivative of f(x) = cosx^2 ---> f ' (x) = 2x(-sinx^2) = -2x sinx^2
yes
so then i take the derivative of -2x sinx^2 ?
yes
okay after i take the derivative of -2x sinx i get -2 cos(2x) so then i take the derivative of -2 cos(2x) right?
use the product rule to get the 2nd derivative : use the formula : f '' = u'v + uv' assuming that u=-2x ---> u' = ... ? v = sinx^2 ---> v ' = ... ?
how would i do it?
can u show it?
do u know derivative of -2x ?
yea
its -2 right?
yes, that's right also, sinx^2 ?
okay hold on
but where does sinx^2 come from i got -2 cos(x) sin(x) ??
for my derivative of cosx^2
f(x) = cosx^2 ---> f ' (x) = 2x(-sinx^2) = -2x sinx^2
okay
it just the first derivative of f(x), now we have to deriv again to get f " then, use the product rule
so now i need to find the derivative of sin^x right? since we found -2x which we got -2
and i got cos (x) sin(x)
so it will be -2 cos(x) sin(x) right?
-2x sinx^2 let u=-2x ---> u' = -2 v = sinx^2 ----> v ' = 2xcoscx^2
sorry, it should v' = 2xcosx^2
okay so how would i set it up in the formula ?
f " = u'v + uv' f " = (-2)sinx^2 + (-2x)2xcosx^2 = -2sinx^2 - 4x^2cosx^2
ok
so what do i do next?
nothing, it is clear :)
okay so the f'''(x) will be -2sinx^2 = 4x^2 cosx^2 ?
i meant f''(x)
are you there?
@RadEn so thats the final answer?

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