differentiation

[gilo26]

Hello all, :wave:

Am struggling to get to grips with some problems set to me.
Any help would be great!! :thumb:

1) Differentiate with respect to x:

a) y = 5 / 3sqrt x

b) y = 1/2 sin x + 1/6cos 3x

c) y = x^2 + ln2x+e^2x

2) Differentiate the following functions:

a) y = (3x-2x^2)(5 +4x)
b) y = 5xsinx
c) y = xe^x/2

Thanks in advance :afrog:

Giles

[azteched]

Things to know:

-The chain rule - look it up
-Product rule - look that up too
-d/dx(sin(x)) = cos(x)
-d/dx(cos(x)) = -sin(x)
-d/dx(ln(x)) = 1/x
-d/dx(exp(x)) = exp(x)

That's all you need to do those.

[beros87]

Things to know:

-The chain rule - look it up
-Product rule - look that up too
-d/dx(sin(x)) = cos(x)
-d/dx(cos(x)) = -sin(x)
-d/dx(ln(x)) = 1/x
-d/dx(exp(x)) = exp(x)

That's all you need to do those.

for your question

1) Differentiate with respect to x:

a) y = 5 / 3sqrt x
d(5/3sqrt x)=5/3(1/x*sqrtx) dx
b) y = 1/2 sin x + 1/6cos 3x
dy= (1/2cos x -1/6 sinx)dx
c) y = x^2 + ln2x+e^2x
dy=(2x + 1/x + e^2x) dx
2) Differentiate the following functions:

a) y = (3x-2x^2)(5 +4x)
dy =[ (5+4x)(-4x+3) + (3x-2x^2)(4)]dx
b) y = 5xsinx
dy=(5sinx +5xcosx)dx
c) y = xe^x/2
dy=(e^x/2+xe^x/2)dx

[gilo26]

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!!

[sunburnt]

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!

[gilo26]

Many thanks, I do believe things are beginning to make sense :bigyello: