Am I applying the chain rule correctly?

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Moved from a technical math section, so missing the template
So i have an equation problem that i need to find the 2nd derivative of, but my understanding of the chain rule is not the best. I tried working it out but i don't know if i did it correctly.

i was given the equation y=4(x2+5x)3

So to take the first derivative, i started off by using the chain rule as i understand it and simplified it to:

y'=[12(x2+5x)2][2x+5]

This is where i start to get confused. I don't really know for certain how I am supposed to take the 2nd derivative of that. I assumed that i needed to actually multiply and simplify the first derivative to take the 2nd, but i don't know if what i did or what i got was correct.

I simplified the first derivative and ended up with: y'=24x5+60x4+1200x3+3000x2

Then i calculated the 2nd derivative and got: y''=120x4+240x3+3600x2+6000x

Would this be correct?
 
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Why not expand the original expression for y and then differentiate it twice?
 
PeroK said:
Why not expand the original expression for y and then differentiate it twice?
Well i thought about doing that but i was unsure whether or not my professor wanted us to use the chain rule. But if he did wanted us to use the chain rule, would it even be feasible to use it on the second derivative?

Is the answer i got correct though at least?
 
To differentiate y' you'd need the chain rule and the product rule.

If you check your answer by doing it the way I suggest, you'll find you made a mistake somewhere.

You should have posted tho in the homework section.
 
Well, you simplified your first derivative wrong. Check your coefficients for x^4 and x^2 terms. It is easier to go the chain rule+product rule, imo
 
P51Mustang said:
Well i thought about doing that but i was unsure whether or not my professor wanted us to use the chain rule.
That possible requirement notwithstanding, it would be a good idea to expand the original function, take its derivative twice, and compare the answer for y'' with what you get using the chain rule.
P51Mustang said:
But if he did wanted us to use the chain rule, would it even be feasible to use it on the second derivative?

Is the answer i got correct though at least?
See above on how you could check.
 
Thanks for the suggestions! I expanded the original equation and recalculated the second derivative and got: y''=120x4+1200x3+3600x2+3000x

But romsofia suggested that my original x4 term's coefficient was wrong, but even after i went back and recalculated carefully, i ended up with 120x4 again.

I just keep getting the wrong answer it would seem. Would it be possible for you to explain or show how to apply the chain and product rule to this problem. I looked back in my calculus book for the chain/product rule, and tried my best to apply it to this problem, but i keep getting answers that are more than likely wrong.
 
P51Mustang said:
Thanks for the suggestions! I expanded the original equation and recalculated the second derivative and got: y''=120x4+1200x3+3600x2+3000x

But romsofia suggested that my original x4 term's coefficient was wrong, but even after i went back and recalculated carefully, i ended up with 120x4 again.

I just keep getting the wrong answer it would seem. Would it be possible for you to explain or show how to apply the chain and product rule to this problem. I looked back in my calculus book for the chain/product rule, and tried my best to apply it to this problem, but i keep getting answers that are more than likely wrong.

Your recalculation is correct. Now could you post an example of how you are getting the wrong answer using the product rule and the chain rule?
 
Dick said:
Your recalculation is correct. Now could you post an example of how you are getting the wrong answer using the product rule and the chain rule?
Thank you for verifying my answer, i was doubting everything i did and i wasnt sure i had done it correctly. But I found my mistake with the chain/product method, though it actually took me way longer to use the chain/product rules. I applied the chain rule and got the first derivative as: 4(x2+5x)2(2x+5)

Then, I used the product rule to try and find the second derivative. But when i went to apply the product rule, when i multiplied the second function by the derivative of the first, i forgot to notice that the first function still has a power in it. So i was getting stuck applying the product rule because i forgot to apply the chain rule a second time to get the derivative of the first function.

This one problem had me ripping my hair out practically but now that its over i don't think ill ever forget the chain/product rule again hahaha.
 
  • #10
P51Mustang said:
Thank you for verifying my answer, i was doubting everything i did and i wasnt sure i had done it correctly. But I found my mistake with the chain/product method, though it actually took me way longer to use the chain/product rules. I applied the chain rule and got the first derivative as: 4(x2+5x)2(2x+5)
@P51Mustang - This is still wrong. Note that you could write your given equation as y = 4u3, where u = x2 + 5x.
Using the chain rule, we have dy/dx = dy/du * du/dx.
P51Mustang said:
Then, I used the product rule to try and find the second derivative. But when i went to apply the product rule, when i multiplied the second function by the derivative of the first, i forgot to notice that the first function still has a power in it. So i was getting stuck applying the product rule because i forgot to apply the chain rule a second time to get the derivative of the first function.

This one problem had me ripping my hair out practically but now that its over i don't think ill ever forget the chain/product rule again hahaha.
 
  • #11
I was talking about when you simplified your first derivative, to check those coefficients, but it seems like you got it.

When I said chain rule+product rule, you need to apply the chain rule WITHIN the product rule because the product rule is f'g+fg', right? So in order to differentiate your first function, you need to apply the chain rule to the 12(x^2+5x)^2 term, which would be your f'.
 

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