Вычислить производную функции.

Question 1

Question 2

Решите задачу:

$1)\ f(x)=x^{10}+x^8+x^3,\\f'(x)=\left(x^{10}+x^8+x^3\right)'=\left(x^{10}\right)'+\left(x^8\right)'+\left(x^3\right)'=\\=10x^{10-1}+8x^{8-1}+3x^{3-1}=10x^9+8x^7+3x^2;\\\\2)\ f(x)=7x^5-3x^9+4,\\f'(x)=\left(7x^5-3x^9+4\right)'=\left(7x^5\right)'-\left(3x^9\right)'+4'=\\=7\left(x^5\right)'-3\left(x^9\right)'+0=7\cdot4x^{5-1}-3\cdot9x^{9-1}=28x^4-27x^8;\\\\3)\ f(x)=4\sqrt x+\frac{x}{3}+\frac{5}{x},$
$f'(x)=\left(4\sqrt x+\frac{x}{3}+\frac{5}{x}\right)'=\left(4\sqrt {x}\right)'+\left(\frac{x}{3}\right)'+\left(\frac{5}{x}\right)'=\\\\=4\left(x^{\frac{1}{2}}\right)'+\frac{x'\cdot3-x\cdot3'}{3^2}+\frac{5'x-5x'}{x^2}=4\cdot\frac{1}{2}\cdot x^{\frac{1}{2}-1}+\frac{1\cdot3-x\cdot0}{9}+\frac{0\cdot x-5\cdot1}{x^2}=\\\\=2x^{-\frac{1}{2}}+\frac{3}{9}+\frac{-5}{x^2}=\frac{2}{\sqrt{x}}+\frac{1}{3}-\frac{5}{x^2};$

$4)\ f(x)=\frac{4}{x}-\frac{\sqrt x}{4}=4\cdot\frac{1}{x}-\frac{1}{4}\cdot\sqrt{x}=4x^{-1}-\frac{1}{4}\cdot\sqrt{x},\\\\f'(x)=\left(4x^{-1}-\frac{1}{4}\cdot\sqrt{x}\right)'=\left(4x^{-1}\right)'-\left(\frac{1}{4}\cdot\sqrt{x}\right)'=\\\\=4\left(x^{-1}\right)'-\frac{1}{4}\left(\sqrt{x}\right)'=4\cdot(-1)x^{-1-1}-\frac{1}{4}\left(x^{\frac{1}{2}}\right)'=\\\\=-4x^{-2}-\frac{1}{4}\cdot\frac{1}{2}x^{\frac{1}{2}-1}=-\frac{4}{x^2}-\frac{1}{8}x^{-\frac{1}{2}}=-\frac{4}{x^2}-\frac{1}{8\sqrt{x}}.$