Question

f(x)=2√x *(x+1)
f`(4)=

f(x)=x+1/x^2
f`(2)=

Answer 1

1)f'(x)=(2/(2√x))*(x+1)+2√x=(1/√x)*(x+1)+2√x;

f'(4)=(1/√4)*(4+1)+2√4=2.5+4=6.5

2)f'(x)=x+(1/x²)=1-2/х³

f'(2)=1-0.25=0.75

Если второе условие такое, f(x)=(x+1)/x², то

f'(x)=((x+1)/x²)'=(x²-2x*(x+1))/x⁴=(-x²-2x)/x⁴

f'(2)=(-4-4)/16=-0.5

Answer 2

$1)\ \ f(x)=2\sqrt{x}\cdot (x+1)\ \ \to \ \ f(x)=2x^{3/2}+2\sqrt{x}\\\\f'(x)=2\cdot \dfrac{3}{2}\, x^{1/2}+2\cdot \dfrac{1}{2\sqrt{x}}=3\sqrt{x}}+\dfrac{1}{\sqrt{x}}\\\\f'(4)=3\cdot 2+\dfrac{1}{2}=6,5$

$2)\ \ a)\ \ f(x)=x+\dfrac{1}{x^2}\ \ ,\qquad \ \ \Big(\dfrac{C}{u}\Big)'=\dfrac{-C\cdot u'}{u^2}\ ,\ \ C=const\\\\f'(x)=1+\dfrac{-2x}{x^4}=1-\dfrac{2}{x^3}\\\\f'(2)=1-\dfrac{2}{8}=1-0,25=0,75\\\\b)\ \ f(x)=\dfrac{x+1}{x^2}\ \ ,\ \ f(x)=\dfrac{1}{x}+\dfrac{1}{x^2}\\\\f'(x)=\dfrac{-1}{x^2}+\dfrac{-2x}{x^4}=-\dfrac{1}{x^2}-\dfrac{2}{x^3}\\\\f'(2)=-\dfrac{1}{4}-\dfrac{2}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}$