Find the derivative of the function.f(x)=sin⁡5(2x3+1)f(x)=\sin^5(2x^3+1)f(x)=sin5(2x3+1)

30 x 2 cos ⁡ ⁡ ( 2 x 3 + 1 ) sin ⁡ 4 ⁡ ( 2 x 3 + 1 ) 30x^2\cos⁡(2x^3+1)\sin^4⁡(2x^3+1) 30 x 2 cos ⁡ ( 2 x 3 + 1 ) sin 4 ⁡ ( 2 x 3 + 1 )

Find the derivative of the function. f ( x ) = sin ⁡ 5 ( 2 x 3 + 1 ) f(x)=\sin^5(2x^3+1) f ( x ) = sin 5 ( 2 x 3 + 1 )

In this problem, we're asked to find the derivative of the function f of x equals sine the power of 5 of 2x cubed plus 1. Now here, since I have a trig function raised to a power, it's gonna make this a bit more clear to go ahead and move that power to the outside so that we can see how exactly we're going to use the chain rule. So rewriting this, I have f of x equals sign of 2x cubed plus 1, all of that raised to that power 5, having pulled that 5 to the outside. Now it's gonna be a bit more clear to see what function is inside and how that looks moving outward. So looking at that innermost function, I have 2x cubed plus 1. Then as we move out, I have sine, and then my outermost function is all of that stuff that's raised to the power of 5. So to find this derivative, we need to use the chain rule, and we're gonna apply it multiple times by continue to multiply by each derivative. So to find the derivative here f prime of x, we're starting from the outside, working our way in. Our outermost function is that step raised to the power of 5. So I'm gonna go ahead and use the power rule here, pulling that 5 out to the front, decreasing that power by 1. Remember that whenever we're working with one function, whatever is inside it remains the same. So I have that 5 on the outside. This is multiplying all of that stuff. Sine of 2x cubed plus 1, all of that remains the same. And then all of that is raised to the power of 5 minus 1, which is 4. So I've taken my first derivative here. We need to continue working our way inside. So my next function here, working my way inside, is the sine function. Now what is the derivative of sine? Derivative of sine is cosine. What is the argument that goes inside of this cosine function? Remember that if we're working in a function, whatever is inside of that function remains the same. So this 2x cubed plus 1 still remains the same here inside of that argument. So I've multiplied by that second derivative here, working my way inside. Now working my way to my last function, my innermost function, this 2x cubed plus 1, I need to multiply by that derivative here. So the derivative of 2x cubed plus 1 using the power rule pulling that 3 out to the front, decreasing that power by 1 gives me 6x. The derivative of a constant one is just 0, so I don't have to worry about that. So we have fully taken our derivative using the chain rule. And now we wanna simplify this a little bit and move some stuff around. So I'm gonna go ahead and move this 6x up to the front, multiply it by that 5. This will end up giving me 30x. So I've accounted for this 5 and this 6x here. And then this is multiplying cosine of 2x cubed plus 1. And then finally, since I have this trig function raised to a power, I'm going to write that last. I'm going to move that 4 back inside to match how our original function was written. And this is going to be sine to the power of 4 of 2x cubed plus 1. And we are all done here. This is our final answer for that derivative. Now feel free to let me know if you have any questions. Remember that the best way to get better at doing these problems is repetition. So I will see you in that next practice problem.