Question

знайти похідну f(x)=1/(3+√х)​

Answer 1

Ответ:

$f'(x)= \frac{ - 1}{2 \sqrt{x} \big(3+\sqrt{x } \: \big)^{2}}$

Пошаговое объяснение:

$f(x) = \frac{1}{3+\sqrt{x}}\\ f(x) = u(v(x))$

где

$v(x) = 3+\sqrt{x}\\ u(v) = \frac{1}{v}$

$f(x) = \frac{1}{3+\sqrt{x}}\\ f(x) = u(v(x)) \\ f'(x) = \left(\frac{1}{3+\sqrt{x}}\right)' = \left(\left(3+\sqrt{x}\right)^{-1}\right)' = \\ = \frac{ - 1}{\big(3+\sqrt{x } \: \big)^{2}} \cdot\left(3+\sqrt{x}\right)' = \\ = \frac{ - 1}{\big(3+\sqrt{x } \: \big)^{2}} \cdot\left(0+ \frac{1}{2 \sqrt{x}}\right) = \\ = \frac{ - 1}{2 \sqrt{x} \big(3+\sqrt{x } \: \big)^{2}}$

Answer 2

Ответ:

$\displaystyle f(x)=\frac{1}{3+\sqrt{x}}\ \ ,\\\\\\\star \ \ \Big(\frac{1}{v}\Big)'=-\frac{v'}{v^2}\ \ ,\ \ v=3+\sqrt{x}\ \ ,\ \ \ \ \ (\sqrt{x})'=\frac{1}{2\sqrt{x}}\ \ \star \\\\\\f'(x)=-\frac{(3+\sqrt{x})'}{(\, 3+\sqrt{x}\ )^2}=-\frac{\dfrac{1}{2\sqrt{x}}}{(3+\sqrt{x})^2}=-\frac{1}{2\sqrt{x}\cdot (3+\sqrt{x})^2}$  

$P.S.\ \ \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}\ \ \ \Rightarrow \ \ \ \Big(\dfrac{1}{v}\Big)'=\dfrac{0\cdot v-1\cdot v'}{v^2}=-\dfrac{v'}{v^2}$