#1
so i got two problems im a little stuck on for some reason. im to find the derivatives of each of the following:

1. y=[(x^2)-3x]cos(2x+10)
i only got to [(x^2)-3x][-sin(2x+1)] + [2x-3][cos(2x+1)] before completely forgetting what i had to do next


2. y= [sqrt(x)] / (x - pi)
and here i just couldnt think of wtf to do.


ty in advance
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#3
Quote by rockon1824
I'm really tired and don't feel like working them out right now, but you need to use the quotient rule on the second one.

yeah i know that, but if i take the derivative of (x - pi) i would just get (1 - 1) no? and that just doesnt seem right to me
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#5
isnt the derivative of (x - pi) just 1.... the pi becomes 0

EDIT for spelling
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#6
yeah alright that makes some sense now, i just kept thinking of pi as a variable that would go to 1 just like x would.
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#8
Quote by shadowsoldier08
so i got two problems im a little stuck on for some reason. im to find the derivatives of each of the following:

1. y=[(x^2)-3x]cos(2x+10)
i only got to [(x^2)-3x][-sin(2x+1)] + [2x-3][cos(2x+1)] before completely forgetting what i had to do next


2. y= [sqrt(x)] / (x - pi)
and here i just couldnt think of wtf to do.


ty in advance


1.- you are missing that dx/dy cos = -sin u DU so you are missing the derivate of the angle of cos
[(x^2)-3x][-sin(2x+10 )*2 ] + [2x-3][cos(2x+10 )]
Thats everything, just multiply everything, which im too lazy to do it in text

2.-
[[(x-pi)(1/2sqrt(x)) ] - [(sqrt(x))(1)]]/(x-pi) ^2
thats your derivate, if you want (your teacher surely does) simplify