Question

Найти производную сложной функции. Даю 15 баллов.
4) f(x) = e^(3x^2-4x+1), x0 = 1
5) f(x) = sin^(2)x, x0 = П/4
6) f(x) = $\sqrt{5x+1}$ , x0 =3

Answer 1

Ответ:

4) 2,  5) 1, 6) 5/8

Объяснение:

4)

$f'(x) = e^{3x^{2} -4x + 1} * (3x^{2} -4x + 1) ' = e^{3x^{2} -4x + 1}* (6x -4)$

$f'(1)=e^{3-4+1}* (6 -4) = 2$

5)

$f'(x) = 2sinx* (sinx)' = 2sinx*cosx=sin2x$

$f'(\frac{\pi}{4} ) = sin\frac{\pi}{2} =1$

6)

$f'(x) = \frac{1}{2\sqrt{5x+1} }*(5x + 1)' = \frac{5}{2\sqrt{5x+1}} \\f'(3) = \frac{5}{2\sqrt{5* 3+1}} = \frac{5}{8}$

Answer 2

$4) \ f(x)=e^{3x^{2}-4x+1 } \\\\f'(x)=(e^{3x^{2}-4x+1 })'=e^{3x^{2}-4x+1 }\cdot(3x^{2}-4x+1)'=\boxed{e^{3x^{2}-4x+1 }\cdot(6x-4)} \\\\f'(x_{0})=f'(1)=e^{3\cdot1^{2}-4\cdot 1+1 }\cdot(6\cdot 1-4)=e^{0}\cdot 2=1\cdot 2=\boxed2$

$5) \ f(x)=Sin^{2}x\\\\f'(x)=(Sin^{2}x)'=2Sinx\cdot(Sinx)'=2Sinx\cdot Cosx=Sin2x\\\\f'(x_{0})=f'\Big(\dfrac{\pi }{4}\Big)=Sin\Big(2\cdot\dfrac{\pi }{4}\Big)=Sin\dfrac{\pi }{2} =\boxed1$

$6) \ f(x)=\sqrt{5x+1} \\\\f'(x)=(\sqrt{5x+1})'=\dfrac{1}{2\sqrt{5x+1}} \cdot(5x+1)'=\dfrac{5}{2\sqrt{5x+1}}=\boxed{\dfrac{2,5}{\sqrt{5x+1}} } \\\\f'(x_{0})=f'(3)=\dfrac{2,5}{\sqrt{5\cdot3+1}}=\dfrac{2,5}{\sqrt{16} } =\dfrac{2,5}{4}=\boxed{0,625}$