1. If f(x)=tan3x, find f'(x) and f'(3).
2. let f(x)=3cos(cosx), find f'(x).
i got stuck. i don't know im doing it right way, f'(x)=tan3x*1
f(x) = tan(3x)
use the chain rule to get
f'(x) = sec^2(3x) * 3 = 3sec^2(3x)
f'(3) = 3sec^2(3)
f(x) = 3cos(cos(x))
again, the chain rule works
let f(u) = 3cos(u) and u(x) = cos(x)
df/dx = df/du * du/dx
= -3sin(u) * -sin(x)
= 3sin(x)sin(cos(x))
i don't get the part sec^2<---why use square?
To find the derivative of a function, you can use the chain rule, which states that if you have a composition of functions, the derivative of the composition is the derivative of the outer function multiplied by the derivative of the inner function.
1. f(x) = tan(3x)
To find f'(x), we can apply the chain rule. The derivative of tan(u) is sec^2(u) * u', where u is the inner function.
- Let u = 3x.
- Find u': derivative of u with respect to x.
- u' = d(3x)/dx = 3
- Find f'(x): f'(x) = d(tan(u))/dx = sec^2(u) * u'.
- f'(x) = sec^2(3x) * 3
To find f'(3), you need to substitute x = 3 into the derivative function.
- f'(x) = sec^2(3x) * 3
- f'(3) = sec^2(3*3) * 3
- Compute the value of sec^2(9) and multiply it by 3 to get the final result.
2. f(x) = 3cos(cos(x))
Again, we will use the chain rule to find the derivative of f(x).
- Let u = cos(x).
- Find u': derivative of u with respect to x.
- u' = d(cos(x))/dx = -sin(x)
- Find f'(x): f'(x) = d(3cos(u))/dx = -3sin(u) * u'.
- f'(x) = -3sin(cos(x)) * u'
Therefore, the derivative of f(x) is -3sin(cos(x)) multiplied by the derivative of the inner function.
If you need further assistance or clarification, feel free to ask!