What is the first and second derivatives of "cosx + xsinx2?

Answer 1

Do you mean "cosx + xsinx^2"?
If so,
first derivative = -sinx + sinx^2 + (xcosx^2)2x
= -sinx + sinx^2 + 2x^2cosx^2

second derivative = -cosx + 2xcosx^2 + 4xcosx^2 - (2x^2sinx^2)2x
= -cosx + 6xcosx^2 - 4x^3sinx^2

Answer 2

I think you mean cosx + x sin ^2 x
derivative of cos x = -sin x
derivative if x sin ^2x using multiplication rule and chain rule
= sin^2 x + x 2 sinx cos x
so derivative of
f(x) = cos x+ x sin ^ x
is
f'(x) = -sin x + sin ^2 x + x sin 2x
now its derivative
= f''(x) = - cos x + 2 sin x cos x + sin 2x .1 + x 2 sin 2x cos 2x
= - cos x + 2 sin 2x + x sin 4x

Answer 3

Ok, assuming that the expression is as follows:

- f(x) = cos(x) + x sin^2(x)
- f'(x) = -sin(x) + sin^2(x) + 2xsin(x)cox(x)
- f''(x) = -cos(x) + 2sin(x)cos(x) + 2 sin(x)cos(x) + 2xcos^2(x) -2xsin^2(x)

f''(x) = -cos(x) + 4sin(x)cos(x) +2x(cos^2(x) - sin^2(x))

Good luck!