cos x
square root x
the derivative of this, i just need a worked example, i just keep getting the wrong answer
$\displaystyle \frac{\cos(x)}{\sqrt{x}} = \frac{\cos(x)}{x^{1/2}} = \cos(x) \cdot x^{-1/2}$ 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:
$\displaystyle \frac{\sin(x)}{x^{1/2}} -\frac{\cos(x)}{2 x^{3/2}}$
$\displaystyle \frac{2x\cdot \sin(x)- \cos(x)}{2x^{3/2}}$
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 $\displaystyle y\prime = (-\sqrt{x}\sin{x} - \frac{x^{-1/2}}{2}\cos{x})/{x}$
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 $\displaystyle 2 \sqrt{x}$.
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}} $ = $\displaystyle \frac {-2xsinx-cosx} {2x \sqrt{x}}$