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Solution of second derivative of function in higher mathematics
1.

y' = x^2 (2^x)' + (2^x) * 2x

= x^2 * 2^x * ln2 + (2^x) * 2x

y ' ' =(x^2 * 2^x * LN2+(2^x)* 2x)* LN2+2x(2^x)ln2+2^x * 2

2.

y' = e^x cos^2 x + e^x (-2cosxsinx)

= e^x cos^2 x- 2e^x cosx sinx

y ' ' =(e^x cos^2 x-2e^x cosx sinx)-2e^xcosx sinx-2e^x(cosx sinx)'

=(e^x cos^2 x-2e^x cosx sinx)-2e^xcosx sinx-2e^x(-sin^2 x+cos^2 x)