Thanks...just one more thing though!
Thanks alot guys, sorry I havent thanked you before, I've been away with work. I have a few more that I've done but am not sure if they are right, it would be great if you could have a quick look :rolleyes:
a) y = 2x + 1 / x-6
b) y = sin x / x
c) y = e^2x / x^2+1
Now this is what I've come up with:
a) -11 / x^2 - 36
b) xcosx - sinx / x^2
c) This one is a bit dodgy I'm sure :eek2:
(x^2 + 1)(2e^2x)-(2xe)^2 / (x^2 +1)^2
I also need to determine the derivatives of these
a) y = (x+5)^3
b) y = xsin^2x
c) y = e^ln3x
.......and I dont know how to :cry: :cry: Any help would be fantastic!!
Re: Thanks...just one more thing though!
Quote:
Originally Posted by gilo26
I also need to determine the derivatives of these
a) y = (x+5)^3
b) y = xsin^2x
c) y = e^ln3x
.......and I dont know how to :cry: :cry: Any help would be fantastic!!
Those are pretty easy; you just need to know the chain rule.
y = (x+5)^3
consider it as y = b^3, and differentiate with respect to b:
dy =3b^2 * db
now, what is db? b = x + 5, so the derivative is just one (1*x => 1)
so the answer is y' = 3(x+5)^2.
Basically, you start with the most outside function, and differentiate from the outside in. You don't need to substitute like I did above; that was just to (hoepfully!) make it clearer.
b) y = xsin^2x
this is the product rule with the chain rule: x * sin(x)^2
1* sin(x)^2 + x * 2 * sin(x) * cos(x)
y' = sin(x)^2 + 2x*sin(x)*cos(x)
c) y = e^ln3x
useful property to know: e^x and ln(x) are inverses of eachother:
e^ln(x) = x;
ln(e^x) = x;
so, y = 3x; y' = 3.
HTH!