Finding the derivative using product rule

**EarlyTravel** · Oct 30th 2012, 12:55 PM

Expand x(2x+1)^4, then find the derivative.

How would I find the derivative?

**AZach** · Oct 30th 2012, 01:15 PM

Originally Posted by EarlyTravel

Expand x(2x+1)^4, then find the derivative.

How would I find the derivative?

You have 2 functions, $f(x)$ and $g(x)$ . Let's say $f(x) = x$ and $g(x) = (2x+1)^4$ .

The product rule is $f'(x)g(x) + g'(x)f(x)$

1) $(\frac{d}{dx}x)(2x+1)^4 + (\frac{d}{dx}(2x+1)^4)x$ . Notice that $(\frac{d}{dx}(2x+1)^4)$ requires the Chain Rule.

The Chain Rule is $f(g(x))' = f'(g(x))*g'(x)$ , so let's say $f(x) = x^4$ and $g(x) = 2x +1$ then $(4(2x+1)^3(2))$

Back to the original problem:

2) $1(2x+1)^4 + 4(2x+1)^3(2))x$ .

3) $(2x+1)^4 + 8x(2x+1)^3$ .

**skeeter** · Oct 30th 2012, 03:44 PM

Originally Posted by AZach

Back to the original problem:

2) $1(2x+1)^4 + 4(2x+1)^3(2))x$ .

3) $(2x+1)^4 + 8x(2x+1)^3$ .

one more worthwhile step ...

$(2x+1)^4 + 8x(2x+1)^3 = (2x+1)^3[(2x+1) + 8x] = (2x+1)^3(10x+1)$

... makes finding critical points a bit less work.

**Soroban** · Oct 30th 2012, 04:20 PM

Hello, EarlyTravel!

Expand $f(x) \:=\:x(2x+1)^4$ , then find the derivative.

How would I find the derivative? . ??

I don't understand your question.
. . You don't know how to expand the function?
. . You don't know how to differentiate a polynomial?
. . You don't know the Product Rule?

How about following the directions?

Expand: . $f(x) \:=\:x(2x+1)^4$

. . . . . . . . . . . $=\;x(16x^4 + 32x^3 + 24x^2 + 8x + 1)$

. . . . . . . . . . . $=\;16x^5 + 32x^4 + 24x^3 + 8x^2 + x$

Differentiate: . $f'(x) \;=\;80x^4 + 128x^3 + 72x^2 + 16x + 1$ .[1]

Product Rule:

. $f(x) \:=\:\overbrace{x}^{f(x)}\cdot\overbrace{(2x+1)^4} ^{g(x)}$

$f'(x) \;=\;\overbrace{1}^{f'(x)}\cdot\overbrace{(2x+1)^4 }^{g(x)} + \overbrace{x}^{f(x)}\cdot\overbrace{4(2x+1)^3\cdot 2}^{g'(x)}$

. . . . $=\;(2x+1)^4 + 8x(2x+1)^3$

. . . . $=\;(2x+1)^3\big[(2x+1) + 8x\big]$

. . . . $=\;(2x+1)^3(10x+1)$

which is equal to [1].

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