# Thread: Derivative of a particular function - help

1. ## 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

2. What ideas have you had so far?

3. 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

4. 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?

5. 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...

Kind regards,

Kepler

6. Looks good to me.

7. 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

8. 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.

9. Thanks again - I said it was a dumb question. You were very helpful. Thanks again.

Kind regards,

Kepler

10. You're welcome. Have a good one!