Product/Quotient rule?

**EarlyTravel** · October 30th 2012, 03:33 AM

Given that $x \sqrt{1+2x}$

Show that $\frac{dy}{dx} = (1+3x)/\sqrt {1+2x}$

First, off, do I use the product rule of quotient rule. Second, could you show me how you got the answer? I got loads more but if you show me how to do one, I think I'll pick up on the rest. Any help is really appreciated, I'm so confused.

**Soroban** · October 30th 2012, 05:27 AM

Hello, EarlyTravel!

Given that: . $y \:=\:x \sqrt{1+2x}$

Show that: . $\frac{dy}{dx} \:=\: \frac{1+3x}{\sqrt {1+2x}}$

Since we have a product: . $y \:=\:x\cdot (1 + 2x)^{\frac{1}{2}}$ , of course we use the Product Rule.

Do you know the Product Rule?

If $y \:=\:f(x)\!\cdot\!g(x)$ , then: . $\frac{dy}{dx} \:=\:f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)$

$\text{We have: }\:y \;=\;\underbrace{x}_{f(x)}\cdot\underbrace{(1+2x)^ {\frac{1}{2}}}_{g(x)}$

$\text{Hence: }\:\frac{dy}{dx} \;=\;\underbrace{1}_{f'(x)}\cdot\underbrace{(1+2x) ^{\frac{1}{2}}}_{g(x)} + \underbrace{x}_{f(x)}\cdot\underbrace{\tfrac{1}{2} (1+2x)^{\text{-}\frac{1}{2}}\cdot 2}_{g'(x)}$

. . . . . . . . $=\;(1 +2x)^{\frac{1}{2}} + x(1 + 2x)^{\text{-}\frac{1}{2}}$

. . . . . . . . $=\;\sqrt{1+2x} + \frac{x}{\sqrt{1+2x}}$

. . . . . . . . $=\;\frac{\sqrt{1+2x}}{1}\cdot {\color{blue}\frac{\sqrt{1+2x}}{\sqrt{1+2x}}} + \frac{x}{\sqrt{1+2x}}$

. . . . . . . . $=\;\frac{1+2x}{\sqrt{1+2x}} + \frac{x}{\sqrt{1+2x}}$

. . . . . . . . $=\;\frac{(1+2x) + x}{\sqrt{1+2x}}$

. . . . . . . . $=\;\frac{1+3x}{\sqrt{1+2x}}$

**EarlyTravel** · October 30th 2012, 05:36 AM

Thank you SO much, honestly, you're a life saver.

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