Derivative help?

Question

Im having some confusion with this derivative problems

y=e^2sin3x
y=2^sec3x

Answer 1

y=e^(2sin(3x))
y' = e^(2sin(3x)) * 2cos(3x)*3
= 6cos(3x)e^(2sin(3x))

y = 2^(sec(3x))
= e^( ln (2) * sec(3x) )
y' = e^( ln (2) * sec(3x) ) * 3 *ln(2)tan(3x)sec(3x)
= 2^(sec(3x)) * 3ln(2)tan(3x)sec(3x).

Answer 2

1. Take the derivative of the exponent and keep e the same.

y'=6cos(3x)*e^(2sin(3x))

2. Take ln of both sides then derive.
y'=ln(2)*(3sec(3x)tan(3x))
*(2^(sec(3x))

Don't forget that the derivative of lny is y'/y. When you multiply it back out, make sure it's in terms of x.

Answer 3

a is a number and u is a function.

(a^u)' = (a^u)(u') ln(a)

(1)
y' = (e^2sin3x) (6cos3x) lne = (e^2sin3x) (6cos3x)


(2)
y' = 3(tan3x)(sec3x)(2^sec3x) ln2

Answer 4

First, is it help you want with finding the derivatives of these functions? Second, is your first function y1 = (e^2)*(sin 3x) or y2 = e^(2* sin 3x)? If y1, then you want to observe that e^2 is a constant so that d(y1)/dx = (e^2)*[d(sin 3x)/dx] =
(e^2)*(cos 3x)*[d(3x)/dx] = (e^2)*(cos 3x)*(3). If y2, then
d(y2)/dx = (e^(2*sin 3x))*[d(2*sin 3x)/dx]= (e^(2*sin 3x))*(2*cos 3x)*[d(3x)/dx] = (e^(2*sin 3x))*(2*cos 3x)*3.
For your other problem, use logarithmic differentiation.