Derivative with square root

[AGlas9837]

Once again, I've been given a first derivative involving a square root and I don't understand how to get there. The problem is: (using <> for brackets)

g(x) = x sq root(x+3)

First I rewrote as x (x+3)^1/2

Then, using the product rule I get: x(1/2)(x+3)^-1/2 + (x+3)^1/2 (1)

Here, if I factor out (x+3)^-1/2, I'm left with: (x+3)^-1/2 <(x) (1/2)+ (x+3)>

The answer given is 3x+6/2 sq root(x+3). I understand the denominator but not the numerator.

 

[soroban]

Hello, AGlas9837!

Your Calculus is fine . . . it's your Algebra I . . .

\(\displaystyle g(x) \:= \:x \sqrt{x+3}\)

\(\displaystyle \text{First I rewrote as: }\:g(x)\:=\:x(x+3)^{\frac{1}{2}}\)

\(\displaystyle \text{Then, using the product rule I get: }\;g'(x) \:=\: x\cdot\frac{1}{2}(x+3)^{-\frac{1}{2}} + (x+3)^{\frac{1}{2}}\)

\(\displaystyle \text{Here, if I factor out }(x+3)^{-\frac{1}{2}},\;\text{ I'm left with: }\x+3)^{-\frac{1}{2}}\left[(x)\left(\frac{1}{2}\right)+ (x+3)\right]\)

\(\displaystyle \text{The answer given is: }\;\frac{3x+6}{2\sqrt{x+3}}\)

\(\displaystyle \text{I understand the denominator but not the numerator.}\)

\(\displaystyle \text{You were left with: }\;(x+3)^{-\frac{1}{2}}\left[\frac{1}{2}x + (x+3)\right] \;=\;(x+3)^{-\frac{1}{2}}\left(\frac{3}{2}x + 3\right)\)

\(\displaystyle \text{Factor out }\frac{1}{2}\!:\;\;\;(x+3)^{-\frac{1}{2}}\cdot\frac{1}{2}(3x + 6) \;=\;\frac{3x+6}{2\sqrt{x+3}}\)

 

[tkhunny]

I really don't like it when algebra gets in the way of learning calculus.

\(\displaystyle x*\frac{1}{2}*\frac{1}{\sqrt{x+3}}\;+\;\sqrt{x+3}\)

Why are you "factoring out"? Why not just add the fractions? Common denominator - that sort of thing?

\(\displaystyle x*\frac{1}{2}*\frac{1}{\sqrt{x+3}}\;+\;\sqrt{x+3}*\frac{2\sqrt{x+3}}{2\sqrt{x+3}}\;=\;\frac{x+2(x+3)}{2\sqrt{x+3}}\)

 

[AGlas9837]

Great! Thanks!