Question

срочно!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$f(x)=5\Rightarrow f'(x)=0$

$f(x)=x\Rightarrow f'(x)=1$

$f(x)=x^2\Rightarrow f'(x)=2x$

$f(x)=4x^3\Rightarrow f'(x)=12x^2$

$f(x)=\sqrt{x}\Rightarrow f'(x)=\frac{1}{2\sqrt{x}}$

$f(x)=\sin x\Rightarrow f'(x)=\cos x$

$f(x)=\cos x\Rightarrow f'(x)=-\sin x$

$f(x)=\mathrm{tg}x\Rightarrow f'(x)=\frac{1}{\cos^2x}$

$f(x)=\mathrm{ctg}x\Rightarrow f'(x)=-\frac{1}{\sin^2x}$