Question

how do you compute (d/dx) [cos^2(x) - cos(x^2)] so that it equals -2cos(x) sin(x) + 2xsin(x^2)

when i do it i get zero

Answer 1

What do you get when you chain rule the first one?

Answer 2

i get zero

Answer 3

let cos x = u
then you have u^2, d/dx(u^2) = 2u * u', 2* cosx * - sinx
= -2cosxsinx

Answer 4

can you use the equations please. its difficult to read and understand how you wrote it

Answer 5

2*Cos(x)*(-Sinx) - (-Sin(2x)*2)

Answer 6

\[(\cos ^{2}x - \cos(x ^{2}))'\]
For
\[\cos ^{2}x\]
Let u = cosx,
Then \[\cos ^{2}x\] = \[u ^{2}\]
Since it is a sum/difference of functions, we can take the derivatives seperately,
so
the derivative of \[u ^{2}\] is 2u * u', which is 2cosx * - sinx

Answer 7

\[\cos x ^{2}\] = \[\sin x^{2} * (x^{2})'\]
You use the chain rule again, you take the derivative of the outer function(cosx), then the inner function (x^2)

Answer 8

ok thanks

Answer 9

-sin* my bad