Verified answer

Объяснение:

2.

[tex]y'=(x^2*cosx)'=2x*cosx+x^2*(-sinx)=2x*cosx-x^2*sinx=\\=2*\frac{\pi }{2} *cos\frac{\pi }{2}- (\frac{\pi }{2})^2*sin\frac{\pi }{2} =\pi *0-\frac{\pi ^2}{4}*1=-\frac{\pi ^2}{4}.[/tex]

3.

[tex]y=x^4+x^3-3x\ \ \ \ \ x_0=2\ \ \ \ \ \ y_k=?\\y_k=y(x_0)+y'(x_0)*(x-x_0)\\y(2)=2^4+2^3-3*2=16+8-6=18.\\y'=(x^4+x^3-3x)'=4x^3+3x^2-3\\y'(2)=4*2^3+3*2^2-3=4*8+3*4-3=32+12-3=41.\ \ \ \ \Rightarrow\\y_k=18+41*(x-2)=18+41x-82=41x-64.[/tex]

Ответ: yk=41x-64.