Find the derivatives of the function
k(x)=sqrt(sin(2x))^3
I think I have the answer and was wondering if i had it correct
k'(x)=2sin(x)^3+8sin(x)
I will interpret your question just the way you typed it
k(x) = √ [ (sin(2x) )^3 ] which is
= ( sin(2x) )^(3/2)
k ' (x) = (3/2)( sin(2x))^(1/2) ( cos(2x)) (2)
= 3 cos(2x) √(sin(2x))
ok thank you
To find the derivative of the function k(x) = sqrt(sin(2x))^3, you can use the chain rule.
Let's break it down step by step:
1. Start by rewriting the function k(x) as k(x) = (sin(2x))^(3/2).
2. Apply the chain rule by differentiating the outer function first. The derivative of (sin(2x))^(3/2) with respect to the inner function (2x) is (3/2)(sin(2x))^(1/2) * d(sin(2x))/dx.
3. Now, differentiate the inner function. The derivative of sin(2x) with respect to x is cos(2x) * d(2x)/dx.
4. Simplify the derivatives obtained so far. d(sin(2x))/dx = 2cos(2x), and d(2x)/dx = 2.
5. Substitute the derived values back into the chain rule expression. (3/2)(sin(2x))^(1/2) * 2cos(2x) * 2.
6. Simplify further. 2 * 2 * (3/2)(sin(2x))^(1/2) * cos(2x).
7. Finally, simplify the expression. 4(sin(2x))^(1/2) * cos(2x).
So, the derivative of the function k(x) = sqrt(sin(2x))^3 is k'(x) = 4(sin(2x))^(1/2) * cos(2x).
Please double-check your answer to see if it matches the derived expression above.