Derivative of a particular function - help

• Nov 22nd 2010, 08:21 AM
kepler
Derivative of a particular function - help
Hi,

I have the following function:

Code:

```h = k1 - k2 a = (k1 - x) / h b = (x - k2) / h y = a * k3 + b * k4 + ((a^3 - a) * k5 + (b^3 - b) * k6) * (h^2) / 6```
I need to know the derivative of y regarding x.

Can someone help me out?

Sorry to bother.

Kind regards,

Kepler
• Nov 22nd 2010, 08:44 AM
Ackbeet
What ideas have you had so far?
• Nov 22nd 2010, 08:51 AM
kepler
Hi,

Substitution; then writing the 3th degree equations in extense. Then calculate the derivative. But I'm afraid I might be lost in the process.

Kind regards,

Kepler
• Nov 22nd 2010, 09:26 AM
Ackbeet
You could do it that way. Perhaps a more compact way would be not to substitute, but to regard a = a(x) and b = b(x), and use the chain rule. What does that give you?
• Nov 22nd 2010, 10:02 AM
kepler
Hi again,

I went with my method and it was a mess...
The substitutiom gave:

Code:

`y = ((k1-x)/h)*k3+((x-k2)/h)*k4+((((k1-x)/h)^3-((k1-x)/h))*k5+(((x-k2)/h)^3-((x-k2)/h))*k6)*(h^2)/6`
and I obtained the derivative:

Code:

`y' = ((3*k6-3*k5)*x^2+(6*k1*k5-6*k2*k6)*x+(3*k2^2-h^2)*k6+(h^2-3*k1^2)*k5+6*k4-6*k3)/(6*h)`
Is this right? I'm not sure...(Wondering)

Kind regards,

Kepler
• Nov 22nd 2010, 10:09 AM
Ackbeet
Looks good to me.
• Nov 22nd 2010, 11:16 AM
kepler
Hi again,

Sorry to ask Adrian - probably a dumb question: but did you check? It's very important that the formula is right.

Sorry to bother you.

Kind regards,

Kepler
• Nov 22nd 2010, 11:18 AM
Ackbeet
I wouldn't have said the answer looks right unless I had checked it. So yes, I did check it. However, you can check it, too, using WolframAlpha.
• Nov 22nd 2010, 11:25 AM
kepler
Thanks again - I said it was a dumb question. You were very helpful. Thanks again.

Kind regards,

Kepler
• Nov 22nd 2010, 11:28 AM
Ackbeet
You're welcome. Have a good one!