Question

Помогите решить! Даю 60б

Answer 1

$1)\; \; g(x)=2ctgx\; \; ,\; \; x_0=-\dfrac{\pi}{4}\\\\\\g'(x)=-\dfrac{2}{sin^2x}\; \; ,\; \; \; g'(-\frac{\pi}{4})=-\dfrac{2}{(-\frac{\sqrt2}{2})^2}=-\dfrac{2\cdot 4}{2}=-4$

$2)\; \; f(x)=\sqrt{x}\cdot x^2\; \; ,\; \; \; f(x)=x^{\frac{5}{2} }\\\\f'(x)=\dfrac{5}{2}\cdot x^{ \frac{3}{2}}=\dfrac{5\, \sqrt{x^3}}{2}$

$3)\; \; f(x)=x^2\; \; ,\; \; x_0=-2\\\\f'(x)=2x\; \; ,\; \; k=tg\alpha =f'(-2)=2\cdot (-2)=-4$

Answer 2

1. g'(x)=-2/cos²x

g'(-π/4)=-2/(1/√2)²=-4

2. f'(x)=2x

f'(-2)=-4

3. f'(x)=(√x*x²)'=(x⁵/²)'=(5x³/²)/2=2.5x√x