Question

derivative of y=-2xsin(x^2)

Answer 1

let u = x^2.

Answer 2

I did

Answer 3

ok, so what are you having trouble with?

Answer 4

one min ill write wat I have so far

Answer 5

ok

Answer 6

y=-2xsin u
u=x^2
derivative of (-2xsinu)= \[\frac{ dy }{ du } -2x \times \sin u + -2x \times \frac{ dy }{ du } \sin u\]
derivative of x^2= 2x

Answer 7

so the answer I get is (-2sin(x^2)-2xcos(x^2))(2x)

Answer 8

but the answer in the book is 4x^2cos(x^2)-2sin(x^2)

Answer 9

oops my bad, i thought you were integrating. anyway, i can explain it.

y=-2xsin(x^2)

ok so keep the constant out front

-2[xsin(x^2]

so take a derivative of the inside

-2[x*(cos(x^2))*(2x) + sin(x^2)]

-2[2x^2*cos(x^2) + sin(x^2)]

-4x^2cos(x^2) -2sin(x^2)

Answer 10

thank you :)

Answer 11

you're welcome