Question

Найдите производную функции

Answer 1

1)f'(x)=2x+4$x^{3}$-9$x^{2}$

f'(0)=0

2)f'(x)=$-$$\frac{(2x-1)'(2x+3)-(2x+3)'(2x-1)}{(2x+3)^{2} }$=$\frac{2(2x+3)-2(2x-1)}{(2x+3)^{2} }$=$\frac{8}{(2x+3)^{2} }$

f'(2)=$\frac{8}{49}$

3)f'(x)=($x^{2}$)'$e^{2x}$+($e^{2x}$)'$x^{2}$=2x$e^{2x}$+2$e^{2x}$$x^{2}$=$2xe^{2x}(1+x)$

4)f'(x)=$\frac{1}{x^{2}+4 } 2x$

5)f'(x)=4*2cosx(-sinx)=-4sin(2x)

6)f'(x)=$-$$\frac{1}{\sqrt{1-(\sqrt{x} })^{2}}*\frac{1}{2\sqrt{x}}=-\frac{1}{\sqrt{1-x}}*\frac{1}{2\sqrt{x}}=-\frac{1}{2\sqrt{x(1-x)}}$

f'($\frac{\pi }{4}$=$-\frac{1}{ \sqrt{ \frac{\pi }{4} (1-\frac{\pi }{4})} } =-\frac{4}{\sqrt{ \pi(4-\pi)} }$

Answer 2

1. f'(x)=(х²+х⁴-3х³+2)'=2x+4x³-6x  f'(0)=0

2. f'(x)=(2*(2x+3)-2*(2x-1))/(2x+3)²=8/(2x+3)²; f'(2)=8/(7)²=7/49

3. f'(x)=2х*е²ˣ+х²*2е²ˣ=2хе²ˣ*(1+х)

4. f'(x)=2х/(х²+4)

5.  f'(x)=-8cosxsinx=-4sin2x

6. f'(x)=(-1/(√(1-x))*(1/2√x); f'(π/4)=(-1/√1-π/4)*(1/2*((√π)/2)))=

(-1/√1-π/4)*(1/2*((√π)/2)))=-√π√(4-π)/π√π=-√(4-π)/π.