Derivative of a product of 3 terms

In summary: You can check the answer by solving for y in terms of x, but it's more complicated than what you've done so far.Yes.
  • #1
knockout_artist
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2

Homework Statement



Taking derivative of 3 term product.

## \frac{d}{dx} (3x^3 y^2 y'^2) ##

Homework Equations



I read that
(abc)' = (ab)c' + (bc)a' + (ca)b'

The Attempt at a Solution


## 9x^2(y^2y'^2) + 6x^3yy'^2 + 6x^3y^2y' ##

is this correct ?
 
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  • #2
knockout_artist said:

Homework Statement



Taking derivative of 3 term product.

## \frac{d}{dx} (3x^3 y^2 y'^2) ##

Homework Equations



I read that
(abc)' = (ab)c' + (bc)a' + (ca)b'

The Attempt at a Solution


## 9x^2(y^2y'^2) + 6x^3yy'^2 + 6x^3y^2y' ##

is this correct ?
No.
The first term is correct, but the last two aren't. The middle term arises from ##3x^3 \cdot \frac d {dx} y^2 \cdot y'^2##. You haven't used the chain rule correctly when differentiating ##y^2##. And similarly for the 3rd term. For that 3rd term, you should end up with a factor of y''.
 
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  • #3
No. While the rule is correct (and a direct result from repeatedly applying the rule for the derivative of a product) you are not doing the derivatives correctly. Note that dy/dx = y’(x) etc
 
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  • #4
knockout_artist said:

Homework Statement



Taking derivative of 3 term product.

## \frac{d}{dx} (3x^3 y^2 y'^2) ##

Homework Equations



I read that
(abc)' = (ab)c' + (bc)a' + (ca)b'

The Attempt at a Solution


## 9x^2(y^2y'^2) + 6x^3yy'^2 + 6x^3y^2y' ##

is this correct ?

Two things. First, you need to learn the chain rule. Second, you can check these things by taking an example, say ##y = x^2## or ##y = \sin x##, and checking your formula. If it's wrong for your example function, then it's wrong in general.
 
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  • #5
knockout_artist said:

Homework Statement



Taking derivative of 3 term product.

## \frac{d}{dx} (3x^3 y^2 y'^2) ##

Homework Equations



I read that
(abc)' = (ab)c' + (bc)a' + (ca)b'

The Attempt at a Solution


## 9x^2(y^2y'^2) + 6x^3yy'^2 + 6x^3y^2y' ##

is this correct ?

Your "abc" equation gives
$$ (3 x^3 y^2 y'^2)' = (3 x^3 y^2 ) (y'^2)' + (3x^3 y'^2) (y^2)' + (3 y^2 y'^2) (x^3)'.$$
After that you must be much more careful than you were.
 
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FAQ: Derivative of a product of 3 terms

What is the formula for finding the derivative of a product of 3 terms?

The formula for finding the derivative of a product of 3 terms is (f'g'h)+(f'gh')+(fg'h'), where f, g, and h are functions.

Can the product rule be used to find the derivative of a product of 3 terms?

Yes, the product rule can be used to find the derivative of a product of 3 terms. The formula for the product rule is (f'g)+(fg'), where f and g are functions.

Are there any special cases when using the product rule to find the derivative of a product of 3 terms?

Yes, there are two special cases to consider when using the product rule for finding the derivative of a product of 3 terms. The first is when one of the functions is a constant, and the second is when two of the functions are the same.

How do you apply the product rule to find the derivative of a product of 3 terms?

To apply the product rule, you must first identify the three functions in the product. Then, take the derivative of each individual function and plug them into the formula (f'g)+(fg'). Simplify the expression to find the final derivative.

Can the chain rule be used to find the derivative of a product of 3 terms?

Yes, the chain rule can be used to find the derivative of a product of 3 terms. However, it may be more complicated and time-consuming compared to using the product rule.

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