Question

Найдите угловой коэффициент касательной к графику функции у=3 sin+12x в его точке с абсциссой х0=-п/2

Answer 1

Y=3sinx +12x

k=tgα=y'
y' =(3sinx +12x)' =3cosx+12

При х₀=-π/2
к=3cos(-π/2)+12=3*0+12=12
Ответ: 12.

Answer 2

Finding the Tangent Line's Slope

To find the slope of the tangent line to the graph of the function \( y = 3 \sin(12x) \) at the point with the abscissa \( x_0 = -\frac{\pi}{2} \), we can use the derivative of the function to find the slope of the tangent line at that point.

The derivative of the function \( y = 3 \sin(12x) \) is given by:

\[ y' = 3 \cdot 12 \cdot \cos(12x) = 36 \cos(12x) \]

Now, we can find the slope of the tangent line at the point \( x_0 = -\frac{\pi}{2} \) by substituting \( x_0 \) into the derivative:

\[ y'(-\frac{\pi}{2}) = 36 \cos(12 \times (-\frac{\pi}{2})) \]

Let's calculate the value of \( y'(-\frac{\pi}{2}) \) using the given derivative.

The search results provided by You.com did not contain the specific information needed to calculate the slope of the tangent line at the given point. Therefore, I will calculate the value of \( y'(-\frac{\pi}{2}) \) using the derivative of the function \( y = 3 \sin(12x) \):

\[ y'(-\frac{\pi}{2}) = 36 \cos(-6\pi) = 36 \cos(\pi) = -36 \]

So, the slope of the tangent line to the graph of the function \( y = 3 \sin(12x) \) at the point with the abscissa \( x_0 = -\frac{\pi}{2} \) is -36.