# Use the chain rule to find the derivative of (3*sin x + 1)^3

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For a function `f(x) = g(h(x))` , the chain rule states that the derivative `(d(f(x)))/(dx) = (d(g(h(x))))/(d(h(x)))*(d(h(x)))/(dx)`

To find the derivative of (3*sin x +1)^3, write y = u^3 where u = 3*sin x +1.

`dy/dx = (dy)/(du)*(du)/(dx)`

= 3*u^2*3*cos x

= 3*(3*sin x +1)^2*3*cos x

= 9*(3*sin x +1)^2*cos x

**The derivative of (3*sin x +1)^3 is 9*(3*sin x +1)^2*cos x**

**Sources:**

The chain rule is differentiation step by step and easy to apply:

`d/dx (3*sinx+1)^3=3*(3*sinx +1)^2* d/dx (3*sinx +1)`

`=3*(3*sinx +1)^2*(3*cos x)`

`= 9*(cosx)*(3*sinx+1)^2`

(3*sin x + 1)^3

You first have to do the power rule, moving the exponent to the front, as a coefficient, and then changing the exponent to one number lower than the original. Keep the inside the same.

3(3*sin x + 1)^2

Now you must multiply this by the derivative of the inside terms. You have to first do the product rule to find the derivative of 3*sin x. A rule states that the derivative of sin x is cos x.

3(3*sin x + 1)^2 * (3*cos x + sin x*0)

Now you can take the derivative of 1, which is zero.

3(3*sin x + 1)^2 * (3*cos x + sin x*0) + 0

Simplified, your answer is

**9(cos x)*(3*sin x + 1)^2**