My Math Forum (http://mymathforum.com/math-forums.php)
-   Differential Equations (http://mymathforum.com/differential-equations/)
-   -   differentiation (http://mymathforum.com/differential-equations/337884-differentiation.html)

 jeho December 5th, 2016 08:05 AM

differentiation

help me with the following question

differentiate:

y=cos inverse [(2cosx+3)/(2+3cosx)]

 skeeter December 5th, 2016 12:44 PM

if $y = \cos^{-1}{u}$ where $u$ is a function of $x$, then $\dfrac{dy}{dx} = -\dfrac{\frac{du}{dx}}{\sqrt{1 - u^2}}$

$u = \dfrac{2\cos{x}+3}{2+3\cos{x}}$

$\dfrac{du}{dx} = \dfrac{(2+3\cos{x})(-2\sin{x}) - (2\cos{x}+3)(-3\sin{x})}{(2+3\cos{x})^2}$

or ...

$\dfrac{d}{dx} \bigg[\cos{y} = \dfrac{2\cos{x}+3}{2+3\cos{x}}\bigg]$

$-\sin{y} \dfrac{dy}{dx} = \dfrac{(2+3\cos{x})(-2\sin{x}) - (2\cos{x}+3)(-3\sin{x})}{(2+3\cos{x})^2}$

either way, finish it ...

 jeho December 5th, 2016 03:34 PM

thank you

l now see where l was making a mistake

 All times are GMT -8. The time now is 12:13 PM.