Trig Differentiation

• Feb 27th 2009, 03:13 PM
I Drink and Derive
Trig 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 PM
nmatthies1
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 PM
I 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 PM
lllll
$\displaystyle \frac{\cos(x)}{\sqrt{x}} = \frac{\cos(x)}{x^{1/2}} = \cos(x) \cdot x^{-1/2}$ 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 PM
e^(i*pi)
Quote:

Originally Posted by I Drink and Derive
cos x
square root x

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

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 PM
mr fantastic
Quote:

Originally Posted by nmatthies1
using the quotient rule $\displaystyle \frac {d} {dx} \frac {cosx} {\sqrt{x}} = \frac {-sinx \sqrt{x} - \frac {1} {2} cosx \frac {1} {\sqrt{x}} } {x}$

Quote:

Originally Posted by I 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

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 PM
nmatthies1
Quote:

Originally Posted by nmatthies1
using the quotient rule $\displaystyle \frac {d} {dx} \frac {cosx} {\sqrt{x}} = \frac {-sinx \sqrt{x} - \frac {1} {2} cosx \frac {1} {\sqrt{x}} } {x}$

yes at that point it is just simplifying

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