Can i get with this differentiation problem? i found f'(x) but can't find f'(5).

The answer is approximately 5.08914650686019
Round this value however you need to



Work Shown:

f(x) = tan(5x)
f ' (x) = 5*sec^2(5x)
f ' (5) = 5*sec^2(5*5)
f ' (5) = 5*sec^2(25)
f ' (5) = 5*(1/(cos^2(25)))
f ' (5) = 5/(cos^2(25))
f ' (5) = 5/((cos(25))^2)
f ' (5) = 5/((0.9912028118635)^2) ... see note below
f ' (5) = 5/0.9824830142461
f ' (5) = 5.08914650686019

Note: make sure your calculator is in radian mode

Y= -2x-1 the negative in front of the 2 is because the slope is negative and the -1 is because it’s down 1 on the y axis.

5because it isyeah i know it is