derivative of y = 2/3sin^(3/2)(x) - 2/7sin^(7/2)(x)

[charnitr]

y= 2/3sin[sup:19c5vb9a]3/2[/sup:19c5vb9a](x) - 2/7sin[sup:19c5vb9a]7/2[/sup:19c5vb9a](x)

 

[Deleted member 4993]

Re: derivative

y= 2/3sin[sup:2ir7vmjj]3/2[/sup:2ir7vmjj](x) - 2/7sin[sup:2ir7vmjj]7/2[/sup:2ir7vmjj](x)

Please show us your work, indicating exactly where you are stuck, so that we know where to begin to help you.

Hint:

\(\displaystyle \frac{dx^n}{dx} \, = \, n\cdot x^{n-1}\)

along with chain rule.

 

[charnitr]

Re: derivative

i got up to where you do:
3/2*2/3sinX[sup:zynx123o]-1/2[/sup:zynx123o] - 7/2*2/7sinX[sup:zynx123o]5/2[/sup:zynx123o]

then they cancel out and i thought it would go sin x - sin x = 0,
but the answer in the back of the book says :
sin[sup:zynx123o]1/2[/sup:zynx123o]xcosx-sin[sup:zynx123o]5/2[/sup:zynx123o]xcosx= cos3x(square root of sinx)

 

[fasteddie65]

Re: derivative

y = (2/3)sin^(3/2) x - (2/7) sin^(7/2) x
y' = (2/3)(1/2) sin^(1/2) x cos x - (2/7)(7/2) sin^(5/2) x cos x =
(1/3) sin^(1/2) cos x - sin^(5/2) x cos x

 

[Deleted member 4993]

Re: derivative

y = (2/3)sin^(3/2) x - (2/7) sin^(7/2) x
y' = (2/3)(1/2) sin^(1/2) x cos x - (2/7)(7/2) sin^(5/2) x cos x =<<<< This is not correct

y' =(2/3)* (3/2)*[sin(x)][sup:6du7pt81](3/2 - 1)[/sup:6du7pt81] * Cos(x) - (7/2)*(2/7)* [sin(x)][sup:6du7pt81](7/2 - 1)[/sup:6du7pt81] * Cos(x)

y' = {sin(x)}[sup:6du7pt81]1/2[/sup:6du7pt81] * Cos(x) - {sin(x)}[sup:6du7pt81]5/2[/sup:6du7pt81] * Cos(x)

y' = {sin(x)}[sup:6du7pt81]1/2[/sup:6du7pt81] * Cos(x) [1- {sin(x)}[sup:6du7pt81]2[/sup:6du7pt81] ]

Finish it now.....

(1/3) sin^(1/2) cos x - sin^(5/2) x cos x