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Quotient Rule help

Differentiate xx+3 \frac{x}{\sqrt{x+3}} simplifying your answer as far as possible.

I've gotten a little stuck.
Here is the solution:

Spoiler



my working is as follows:
let u=x u = x and v=x+3 v = \sqrt{x+3}

dudx=1\frac{du}{dx} = 1 and dvdx=12(x+3)−12 \frac{dv}{dx} = \frac{1}{2}(x+3)^\frac{-1}{2}

thus using quotient rule: x2x+3−x+3x+32=x2x+3−x+3x+3 \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{\sqrt{x+3}^2} = \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{x+3}

Where do I go from here? :redface:
Reply 1
Avatar for ttoby
ttoby
15
Original post by purplefrog
Differentiate xx+3 \frac{x}{\sqrt{x+3}} simplifying your answer as far as possible.

I've gotten a little stuck.
Here is the solution:

Spoiler



my working is as follows:
let u=x u = x and v=x+3 v = \sqrt{x+3}

dudx=1\frac{du}{dx} = 1 and dvdx=12(x+3)−12 \frac{dv}{dx} = \frac{1}{2}(x+3)^\frac{-1}{2}

thus using quotient rule: x2x+3−x+3x+32=x2x+3−x+3x+3 \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{\sqrt{x+3}^2} = \frac{\frac{x}{2\sqrt{x+3}} - \sqrt{x+3}}{x+3}

Where do I go from here? :redface:


You made a small mistake with which way round the quotient rule goes. You should have got x+3−x2x+3x+32 \frac{\sqrt{x+3} -\frac{x}{2\sqrt{x+3}} }{\sqrt{x+3}^2}. Do that same simplification step you did before, then multiply top and bottom of the fraction by x+3\sqrt{x+3}.
Reply 2
Avatar for haloturk
haloturk
11
your numerator is the wrongway round .... Wherever v is it goest before the U..
can't really explain it but it should be Root(x+3)-(x/2root(x+3) and then divide by v^2

^^ ahh he explained it clearer then i ever could have :smile:
(edited 13 years ago)
If you hate the quotient rule more than the product rule, then you could always just use the product rule. A question will never specifically ask you to use one method over the other.

For me, turning xx+3 \frac{x}{\sqrt{x+3}} into x(x+3)−12x(x+3)^{-\frac{1}{2}} and using the product (and chain) rule is simpler and faster than trying to remember the quotient rule.

Just an idea though. Use whatever you feel more comfortable with.
Reply 4
Avatar for alexs2602
alexs2602
22
Original post by purplefrog
x

I'm going to start from scratch so it makes sense to me

∫xx+3dx=∫uv=v⋅dudx−u⋅dvdxv2 \int \frac {x}{\sqrt{x+3}}dx = \int \frac {u}{v} = \frac {v \cdot \frac {du}{dx} - u \cdot \frac {dv}{dx}}{v^2}

u=x    v=x+3=(x+3)12 u = x \ \ \ \ v = \sqrt{x+3} = {(x+3)}^{\frac {1}{2}}

dudx=1     dvdx=(x+3)−122=12x+3 \frac {du}{dx} = 1 \ \ \ \ \ \frac {dv}{dx} = \frac {{(x+3)}^{- \frac {1}{2}}}{2} = \frac {1}{2 \sqrt{x+3}}

∫xx+3dx=x+3−x2x+3x+32 \int \frac {x}{\sqrt{x+3}}dx =\frac {\sqrt{x+3} - \frac {x}{2 \sqrt {x+3}}}{{\sqrt {x+3}}^2}

x+3−x2x+3x+3⋅2x+32x+3=2(x+3)−x2(x+3)(x+3)12\frac {\sqrt{x+3} - \frac {x}{2 \sqrt {x+3}}}{x+3} \cdot \frac {2 \sqrt {x+3}}{2 \sqrt {x+3}} = \frac{ 2(x+3) - x}{2(x+3)(x+3)^{\frac {1}{2}}}

x+62(x+3)32 \frac {x+6}{2(x+3)^{\frac {3}{2}}}
(edited 13 years ago)

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