Question

Найдите производную функции в указанной точке:

а) y=x^4, x=1;

б) y=2cosx+1, x=pi/4

в) y= √x(x+2), x=4

Answer 1

а).

$y = x^4$.

$y'=(x^4)'=4x^3$.

При $x=1$:

$4x^3=4 \cdot 1^3 = 4$.

ответ: $4.$б).

$y = 2 \;\! \cos \;\! x + 1$.

$y' = (2 \;\! \cos \;\! x)' = 2 \cdot (\cos \;\! x)' = -2 \;\! \sin \;\! x$.

При $x= \pi /4$:

$-2 \;\! \sin \;\! x = -2 \;\! \sin \;\! \dfrac{\pi}{4} = -2 \cdot \dfrac{\sqrt{2}}{2} = - \sqrt{2}$.

ответ: $-\sqrt{2}.$в).

$y = \sqrt{x} \cdot (x+2)$.

$\displaystyle y' = (\sqrt{x} \cdot (x+2))' = (x^{1/2} \cdot (x^1+2))' = (x^{3/2} + 2 \cdot x^{1/2})' =\\\\= \frac{3}{2} \cdot x^{1/2} + 2 \cdot \frac{1}{2} \cdot x^{-1/2} = \frac{3 \cdot \sqrt{x} }{2} + \frac{1}{\sqrt{x}} = \dfrac{3x+2}{2 \cdot \sqrt{x} }$

При $x=4$:

  $\displaystyle \frac{3x+2}{2 \cdot \sqrt{x} } = \frac{3 \cdot 4 +2}{2 \cdot \sqrt{4} } =\frac{14}{4} = \frac{7}{2} = 3,5$.

ответ: $3,5 \; .$