Answer to Question #305304 in Calculus for Anniee

Question

Find the first and second derivative of the following:

1. y=cos x²

2.y=sinx cos x²

3.y=xsin x²

4.y=x cos x²

Answer 1

$1) y=\cos x^2\\ y' =-\sin x^2\cdot 2x=-2x\sin x^2\\ y''=-2\sin x^2 - 2x\cdot\cos x^2\cdot2x=\\ =-2\sin x^2-4x^2 \cos x^2;\\ 2) y=\sin x\cos x^2\\ y'=\cos x \cos x^2 - 2x\sin x \sin x^2\\ y''=-\sin x \cos x^2 -\cos x \sin x^2 \cdot 2x -\\ -\cos x \sin x^2 \cdot 2x - \sin x\cos x^2\cdot2x\cdot2x-\\ -\sin x \sin x^2 \cdot2=\\ =-\sin x\cos x^2 -4x \cos x \sin x^2-\\ -4x^2 \sin x\cos x^2 -2\sin x \sin x^2;\\ 3) y=x \sin x^2\\ y'=\sin x^2 +x\cdot\cos x^2\cdot2x=\\ =\sin x^2 +2x^2 \cos x^2\\ y''=\cos x^2 \cdot 2x+4x \cos x^2 -2x^2 \sin x^2 \cdot2x=\\ =6x \cos x^2-4x^3 \sin x^2;\\ 4) y=x \cos x^2\\ y'= \cos x^2 -x \sin x^2 \cdot 2x=\\ =\cos x^2 - 2x^2 \sin x^2\\ y''= -\sin x^2\cdot 2x- 4x \sin x^2-2x^2 \cos x^2\cdot2x=\\ =-6x \sin x^2-4x^3\cos x^2.$