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Using the chain rule, how do i solve this problem?
g(x)=(sin3x)^3
g'(x)=??
3sin(3x)^2 • 3
i think
*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question
*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question
*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question
let u=sin^2 (3x)
du= 2sin(3x)*cos(3x)*3
check that.
To find the derivative of g(x) = (sin(3x))^3 using the chain rule, follow these steps:
Step 1: Identify the inner function (u) and its derivative (du).
- Let u = sin^2(3x)
- To find du, use the chain rule for the function u = sin^2(3x):
du = 2sin(3x)cos(3x) * 3
Step 2: Rewrite g(x) in terms of u.
- Since g(x) = (sin(3x))^3, we can rewrite it as g(x) = u^3.
Step 3: Apply the chain rule by differentiating g(x) with respect to u and multiplying by du.
- dg(x)/du = 3u^2
- Multiply dg(x)/du by du to get the derivative in terms of x:
g'(x) = 3u^2 * du
Step 4: Substitute the value of u and du back into the expression.
- g'(x) = 3(sin^2(3x))^2 * 2sin(3x)cos(3x) * 3
Simplifying further, we have:
- g'(x) = 18sin^2(3x)sin(3x)cos(3x)
Therefore, the derivative of g(x) = (sin(3x))^3 is g'(x) = 18sin^2(3x)sin(3x)cos(3x).