cos x

square root x

the derivative of this, i just need a worked example, i just keep getting the wrong answer

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- Feb 27th 2009, 03:13 PMI Drink and DeriveTrig Differentiation
__cos x__

square root x

the derivative of this, i just need a worked example, i just keep getting the wrong answer - Feb 27th 2009, 03:21 PMnmatthies1
using the quotient rule $\displaystyle \frac {d} {dx} \frac {cosx} {\sqrt{x}} = \frac {-sinx \sqrt{x} - \frac {1} {2} cosx \frac {1} {\sqrt{x}} } {x} $

- Feb 27th 2009, 03:24 PMI Drink and Derive
yes i got to there but the answer takes it further, like it has a denominator of 2x ROOT x in the answers, anyway to go furhter than that?

thanks anyway - Feb 27th 2009, 03:28 PMlllll
$\displaystyle \frac{\cos(x)}{\sqrt{x}} = \frac{\cos(x)}{x^{1/2}} = \cos(x) \cdot x^{-1/2}$ which should simplify the calculation

so once we differentiate we get:

$\displaystyle -\sin(x) \cdot x^{-1/2} +\left(-\frac{1}{2}x^{(-1/2-1)} \cdot \cos(x) \right)$

simplifying we have:

$\displaystyle \frac{\sin(x)}{x^{1/2}} -\frac{\cos(x)}{2 x^{3/2}}$

$\displaystyle \frac{2x\cdot \sin(x)- \cos(x)}{2x^{3/2}}$ - Feb 27th 2009, 03:29 PMe^(i*pi)
I would use the quotient rule which for $\displaystyle \frac{u}{v}$ is:

$\displaystyle y\prime = (vu\prime - uv\prime)/v^2$

Let $\displaystyle u = \cos{x}$ and $\displaystyle v = \sqrt{x}$ so that $\displaystyle u\prime = -\sin{x}$ and $\displaystyle v\prime = \frac{1}{2}x^{-1/2} $

Putting these together we get $\displaystyle y\prime = (-\sqrt{x}\sin{x} - \frac{x^{-1/2}}{2}\cos{x})/{x}$ - Feb 27th 2009, 03:29 PMmr fantastic
To save people wasting time re-inventing the wheel, it would help if you showed where you got stuck so the that help could be given from that point.

Multiply the numerator and denominator of this answer by $\displaystyle 2 \sqrt{x}$. - Feb 27th 2009, 03:31 PMnmatthies1
yes at that point it is just simplifying

$\displaystyle \frac {-sinx \sqrt{x} - \frac {1} {2} cosx \frac {1} {\sqrt{x}} } {x}$ = $\displaystyle \frac {-sinx \sqrt{x}} {x} - \frac {\frac {1} {2} cosx \frac {1} {\sqrt{x}}} {x} $ = $\displaystyle \frac {-sinx} {\sqrt{x}} - \frac {cosx} {2x \sqrt{x}} $ = $\displaystyle \frac {-2xsinx-cosx} {2x \sqrt{x}}$