Derivative of a product of 3 terms

[knockout_artist]

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 ?

 

[Mark44]

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''.

 

[Orodruin]

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

 

[PeroK]

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.

 

[Ray Vickson]

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.