Part a of the question asks us to prove that F Times g times h if we take its derivative is equal to F prime G H plus F g Prime H plus F G h Prime. So to do that, we're going to need to use the product rule, which I have written over here in terms of U and V so that we don't get it confused with F and G. So we'll need to identify our you, you Prime V and v prime. So are you, we can say is F and that means that the derivative of you is just the derivative of F and V, We can say is just g h. And now if we want the derivative of G H, we have to use the product rule within the product rule to find the derivative of G. H. So we can say that it is G prime times H plus H Prime times G again just rewriting it in the form of the product rule. And so now we can put these all back together into the original products rule. So we have new prime, which is F prime times V, which is G H plus the prime is G prime H plus H prime G times you which is F So if we distribute the F, we get f prime G H plus half G prime h plus Uh huh G h prime. And now we have proved that this is equal to the derivative of f times g times h. And so now for part B, we want to use that statement to prove this statement. But first, we need to set f equals G and equal to H as the problem tells us So now we can rewrite this function here as simply s cute. Since f times g times H is now f times F 10 f So if we want the derivative cubed, we rewrite f prime G becomes f h becomes f f stays, F g prime becomes f prime. Age becomes f and F becomes f g becomes f and H prime becomes f prime. So now we can rewrite this as three f prime because we have our three F primes here, here and here and f squared. And so now we can rewrite this in a more fancy way, as provided by the problem up here. This simply equals three times the derivative of f times the function f of X squared. So we have shown that this statement provided by B is true. And now finally, we can use this new rule that the derivative of F cubed is equal to three times derivative, the derivative of f times F X squared and we can use that to solve. For this new function y equals x to the fourth plus three x cubed plus 17 explicit 82 cube. So our affects is the inside of the function X to the four plus three X cubed plus 17 X plus 82. And now we want to find what half of X Cube is. So we have X to the fore plus three x cubed plus 17 x plus 82 cubed again. And so we know that this here is equal to this What we've proved in part two so we can say three times the derivative of F. Now, if we want to find the derivative of F, we'll need to use the product rule on each of the terms within ffx since we know it's a polynomial again product rule is the derivative of X to the end is equal to end times X to the end minus one. So first, we'll you will use the product rule on the term X to the four. We see that four is R n So we dropped that in front of X and now four minus one is equal to three. So we add the next term and we can see that our end is equal to three. So we drop that down and multiply it by the coefficient. Three to get nine x and three minus one is equal to X squared. And our next term is 17 x and we can see that r n is simply one. So we drop that down and multiplied by the coefficient. 17 we get 17, one minus one is equal to zero, so we would have X to the zero. But we know that anything raised to the zero power is one. So we just have 17. And for our final term 82 we can see that it is a constant and the derivative of a constant is simply zero. So this here is our derivative of F of X, and so Now we can plug that in to what we got in part B. So three times for X cubed plus nine X squared plus 17. And now we simply have the function f f X squared so we can multiply it. Bye x to the fore. Plus three x cubed plus 17 X plus, 82 squared. And now this is our answer for part C of the problem.