what is derivative of sin^2 (3x-1)^2 ?
dy/dx (sin^2 (x))= 2 sin x cos x
dy/dx (3x-1)^2= 6(3x-1)
chain rule?
= 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1)
Please help. Thank you
I'm not sure if this is correct. Can someone clarify thanks.
you are correct
Or, you could go on and say
sin (2(3x-1)^2) * 6(3x-1)
I got 12 sin (3x-1)^2 cos (3x-1)^2 * (3x-1) as my final answer. Is this in the correct simplified form ?
it is ok, but you can use
2sinu cosu = sin(2u)
to make it what I had:
6(3x-1) sin(2(3x-1)^2)
If you want to retain your original argument of sin and cos, then leave it as you had it.
I have taken that into consideration; thank you!
To find the derivative of the function sin^2((3x-1)^2), you can apply the chain rule. The chain rule states that if you have a composite function, such as f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x). In this case, you have f(x) = sin^2(x) and g(x) = (3x-1)^2.
First, let's find the derivative of f(x) = sin^2(x). The derivative of sin^2(x) is given by:
dy/dx (sin^2 (x)) = 2 sin(x) cos(x)
Next, let's find the derivative of g(x) = (3x-1)^2. The derivative of (3x-1)^2 can be found using the power rule. For a function (ax+b)^n, the derivative is given by:
dy/dx ((ax+b)^n) = n(ax+b)^(n-1) * a
Applying the power rule to (3x-1)^2, we have:
dy/dx ((3x-1)^2) = 2(3x-1)^(2-1) * 3
= 6(3x-1)
Finally, applying the chain rule, we multiply the derivatives of f(g(x)) and g(x):
dy/dx (sin^2 ((3x-1)^2)) = 2 sin ((3x-1)^2) cos ((3x-1)^2) * 6(3x-1)
= 12(3x-1) sin ((3x-1)^2) cos ((3x-1)^2)
So, the derivative of sin^2((3x-1)^2) is 12(3x-1) sin ((3x-1)^2) cos ((3x-1)^2).