Find the derivative of f(x)=3/7sin^2(3x)

Question

I got 9sin6x/7

But a website says 18cos(3x)sin(3x)/7

Which one is it? And how do I get there? I'm a little confused.

Answer 1

Whenever we see an equation that seems like it has a lot going on, it helps to break it down into more manageable bits, making it more obvious which derivation rules need to be used. This doesn't need to be done every time, but it's helpful as you're building your intuition when it comes to derivation.

First, we convert it from single-line to general mathematical notation to make it easier to understand the layers of each operation being performed within the function:

f(x) = 3/7sin^2(3x)

3

f(x) = –– sin²(3x)

7

3

f(x) = –– [sin(3x)]²

7

From here, we start breaking the function down into smaller chunks:

3

f(u) = –– [u]² where u = sin(3x)

7

3

f(u) = –– u² where u = sin(v) and v = 3x

7

If you didn't know already, calculus was developed separately and independently by both Sir Isaac Newton and Gottfried Leibniz. And since then, many mathematicians have put their own flare on how to describe calculus. Because of that, there's different ways to notate the same concepts, including derivatives. We could use Lagrange's notation with the derivative as f'(x), but for this scenario, I think it'll be more intuitive to use Leibniz's notation with the derivative as d(f(x))/dx or:

d(f(x)) dy

––––– = ––

dx dx

Back to where we left off:

3

f(u) = –– u² where u = sin(v) and v = 3x

7

We take the derivative of the main equation with respect to u using the Power Rule:

d(f(u)) 3 6

––––– = –– 2 u^2-1 = –– u

du 7 7

The problem is, we need the derivative with respect to x, but to get to that, we first need the derivative with respect to v. From earlier:

u = sin(v)

du

–– = cos(v)

dv

Which we can rearrange as:

du = cos(v) dv

Plug into the main equation:

d(f(v)) 6

–––––––– = –– sin(v)

cos(v) dv 7

And rearrange to the standard format:

d(f(v)) 6

–––––– = –– sin(v)cos(v)

dv 7

We're getting closer! Let's do this process again, but with respect to x now. From way earlier:

v = 3x

dv

–– = 3

dx

dv = 3dx

Plugging in and simplifying:

d(f(x)) 6

–––––– = –– sin(3x)cos(3x)

3dx 7

d(f(x)) 6

–––––– = 3 –– sin(3x)cos(3x)

dx 7

d(f(x)) 18

–––––– = –– sin(3x)cos(3x)

dx 7

Now that we have our answer, you might be able to see where the "chain" in Chain Rule comes from. We're simply linking together a series of nested derivatives to get to that of our initial equation. If you follow this process enough times, you'll start to see where the shortcuts and general formulas came from in the first place.

Answer 2

Hi Celine,

you see a few methods by dear expert tutor. Now I give you a new form that always works for any trigonometry function.

Y = K sinⁿ ( u (x) ) that k = number n = number u(x) is a kind of function

Y' = K n sin^n-1 ( u (x) ) cos(u(X)) u'(x)

f(x)=3/7sin^2(3x) that K = 3/7 n= 2 and u(x) = 3x

f'(x) = 3/7 * 2 * sin(3x) cos(3x) * 3

f'(x) = 18/7 * sin(3x) cos(3x)

==============================================

for cos to the power n:

y = K cos ⁿ(u(x)) ------> y' = - K * n *cos ^n-1(u(x)) sin(u(x)) u'(x)

I hope it is useful,

Minoo

Answer 3

I think the issue was that you were seeing it as sin(2*3x) and not a sin²(3x). In general, the angle will not change when you get to the answer.

IF the question was 3/7*sin(6x), your answer would be almost be correct, 3/7*6*cos(6x) = 18/7*cos(6x).

Foe the problem give, the outside function is a square and the inside function is sine.

Answer 4

Hey! Lets solve this shall we ?

So, we must find the derivative to the given function f(x) = (3/7)sin²(3x) = (3/7)[sin(3x)]². (Notice that it can be rewritten how i wrote it). We find the derivative of this function using the Power Rule, Chain Rule and Product Rule together such that,

f ' (x) = f(x) * g'(x) + g(x) * f '(x)

Let f(x) = 3/7 and g(x) = [sin(3x)]²

f ' (x) = (3/7) * [2 (sin3x) (cos3x*(3))] + [sin(3x)]² * d/dx (3/7)

f ' (x) = (3/7) * (6) cos(3x)sin(3x) + 0.............because d/dx (3/7) = 0

f ' (x) = (18/7)cos(3x)sin(3x)............After multiplying 6 times 3.

I hope this helped!