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#1 2011-09-30 14:07:52

IMMORTALX
Member
Registered: 2011-09-30
Posts: 4

Calculus: Find the derivative of the function.

1. f(x) =x/√7-3x Note: The square root is over 7-3x.

2. y = xsin1/x

3. f(t) =√t/t^2+4 Note: The square root is over t/t^2+4.

4. y = [x^2+(1-3x)^5]^3.

5. sin(sin(sinx)).

6. y = √x+√x Note: The first square root is over x+√x.

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#2 2011-09-30 15:59:33

reconsideryouranswer
Member
Registered: 2011-05-11
Posts: 171

Re: Calculus: Find the derivative of the function.

IMMORTALX,

you have to post your work thus far (your attempts),
so that you can be helped.  Show us where you are
stuck by showing your work . . . first.

Last edited by reconsideryouranswer (2011-09-30 16:11:37)


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#3 2011-09-30 19:02:02

Bob
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Registered: 2010-06-20
Posts: 10,583

Re: Calculus: Find the derivative of the function.

hi IMMORTALX

On this forum, we try to help people with maths.  But we don't just do their homework for them. 
So, I'll start with your first post and give some hints. 
Either that'll be enough for you to do them or at least you can post back exactly where your difficulty is.

1. f(x) =x/√7-3x Note: The square root is over 7-3x.

So you mean:

You'll need to apply the chain rule to the denominator as part of a quotient rule for the whole thing.
I'd do dv/dx where v = root(7-3x) first.
Then you are ready to do the quotient as you know dv/dx.


2. y = xsin1/x


This also requires the application of one rule (chain) within another (product).

Start with z = sinw where w = 1/x and do the chain rule on this.
Then use the product rule on x.z

I'll stop there for you to try these.

Bob

Last edited by Bob (2011-09-30 19:06:50)


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