What's f'(x) for: f(x) = sin^4 3x?

Answer 1

First off, this problem involves the chain rule. I think the person who said package rule really meant product rule, which is not requried here.

The chain rule allows us to find the deriviative of a function which we don't know how to differentiate in terms of functions which we do know how to differentiate. In this case, we know how to differentiate sin(3x), but not (sin(3x))^4 (I think it is less confusing to write it that way given the lack of mathematical notation available here). The chain rule, in general, says that, if we have two functions, g(x) and h(x), we can find the deriviative of the compostie function g(h(x)) as follows:
(g(h(x))' = g'(h(x)) * h'(x)
So, let us have hour h(x) function defined as:
h(x) = sin(3x)
It's derivative is then:
h'(x) = 3cos(3x)
Now, lets define g(x) as:
g(x) = x^4
g'(x) = 4x^3
g composed on h is then:
g(h(x)) = g(sin(3x)) = (sin(3x))^4
g' composed on h is:
g'(h(x)) = 4(sin(3x))^3
Now we have everything we need to use the chain rule, and find the derivative of g(h(x))
(g(h(x))' = 4(sin(3x))^3 * 3cos(3x)
= 12*cos(3x)*(sin(3x))^3
In practice, one doesn't need to do this process so formally. Any time you have some base function you know how to differentiate raised to some power, to take the derivative you simply bring the power out front, lower the power by one, and mulitply by the derivative of the base.

Answer 2

f'(x) = 4sin^3(3x).cos(3x).3
= 12sin^3(3x).cos(3x)

Don't ask home work questions in this board. Anything that you
can obtain straight from text book, you should not waste others'
time here.

Answer 3

f(x)=Sin^4(3x)
f'(x)=dy/dx sin^4(3x)

Let u equal sin(3x) and f(u) equal u^4

now the equation becomes (4u^3)(dy/dx sin(3x)

then let u2 equal 3x and f(u2) equal sin(u2)

Now the equation becomes (4u^3)(3)(cos(u2))

substitute u and u2 in

(4sin^3(3x)(3)(cos(3x)
=12sin^3(3x)cos3x)

THIS IS CALLED THE CHAIN RULE

Answer 4

12sin^3(3x)cosx
chain rule

do u r hw next time

Answer 5

f(x) = [sin(3x)]^4

f'(x) = 4[sin(3x)]^3 * cos(3x) * 3

f'(x) = 12*[sin(3x)]^3 * cos(3x)

Answer 6

4sin^3 3x 3cos3x + 3sin^4x

( hope you understand your own notation ;)

Answer 7

f'(x)=12cos3xsin^3 3x
f(x) is the same as (sin3x)^4
so you would just use package rule, i think thats what it is called.