Question

(-7tgx)''- производная​

Answer 1

$f(x)=-7tgx\\f'(x)=\frac{-7}{cos^2x}$

Answer 2

To find the second derivative of the function (-7tgx), we'll need to differentiate it twice with respect to x. Let's start by finding the first derivative:

f(x) = -7tgx

Using the chain rule, we can find the first derivative:

f'(x) = (-7)(sec^2x)(1)

Remember that sec^2x is the derivative of tanx.

Now, let's find the second derivative by differentiating f'(x):

f''(x) = [(-7)(sec^2x)]' * (1)

To differentiate (-7)(sec^2x), we'll apply the product rule:

[(-7)(sec^2x)]' = (-7)'(sec^2x) + (-7)(sec^2x)'

The derivative of -7 is 0, so the first term becomes 0.

[(-7)(sec^2x)]' = 0 + (-7)(sec^2x)'

Now, we need to find the derivative of sec^2x:

(sec^2x)' = 2(secx)(secx')(1)

The derivative of secx is secx * tanx. Substituting this in, we get:

(sec^2x)' = 2(secx)(secx * tanx)(1)

Simplifying, we have:

(sec^2x)' = 2sec^2x * tanx

Now, substituting this back into the expression for the second derivative:

f''(x) = [(-7)(sec^2x)]' * (1) = (-7)(sec^2x)' = (-7)(2sec^2x * tanx) = -14sec^2x * tanx

Therefore, the second derivative of the function (-7tgx) is -14sec^2x * tanx.