Product Rule Question

**centenial** · Mar 22nd 2010, 05:02 PM

Hi,

I'm asked to differentiate $f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3$ .

Using the product rule, I know that $f'(x) = u'v + uv'$ .

What I'm confused about, is what to set $u$ and $v$ to in the above function. Would $u=3(x^2 + 2)(4x^2 - 5x^4)$ and $v = -3$ ?

Thanks,

**Archie Meade** · Mar 22nd 2010, 05:38 PM

Originally Posted by centenial

Hi,

I'm asked to differentiate $f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3$ .

Using the product rule, I know that $f'(x) = u'v + uv'$ .

What I'm confused about, is what to set $u$ and $v$ to in the above function. Would $u=3(x^2 + 2)(4x^2 - 5x^4)$ and $v = -3$ ?

Thanks,

No, not at all.

u and v are functions of x.

Since the derivative of a constant is zero, you can forget about the -3 at the end.

Also the multiplier 3 can be used as simply an external factor.

Hence

$u(x)=x^2+2$

$v(x)=4x^2-5x^4$

$\frac{d}{dx}\left[3\left(x^2+2\right)\left(4x^2-5x^4\right)-3\right]=3\frac{d}{dx}\left[\left(x^2+2\right)\left(4x^2-5x^4\right)\right]$

You could of course just multiply out the expression and differentiate term by term also.

**skeeter** · Mar 22nd 2010, 05:39 PM

Originally Posted by centenial

Hi,

I'm asked to differentiate $f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3$ .

Using the product rule, I know that $f'(x) = u'v + uv'$ .

What I'm confused about, is what to set $u$ and $v$ to in the above function. Would $u=3(x^2 + 2)(4x^2 - 5x^4)$ and $v = -3$ ?

Thanks,

product rule ...

$u = 3(x^2+2)$

$v = (4x^2-5x^4)$

the -3 term at the end of the expression is inconsequential since the derivative of -3 is 0

**centenial** · Mar 22nd 2010, 05:42 PM

Thanks! I think I understand now.

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