Derivative of polynomial function: P_3(x) = (9f(x_4) - 27f(x_3) + 27f(x_2) - 9f(x_1))/(2b^3) *...

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Tuta

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Hi everybody,

I need to find the formula for P'(a) but I don't know how to derive this third-degree polynomial function:

1pwFnDV.png


a and b are constants but those f(x1)...f(x4) confuse me a lot. Any ideas?
 
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f(x_1) is the value of the function f(x) when x=x_1. That is, f(x_1) is a constant, just like f(5) is a constant.

Here is a helpful hint. [f(x)g(x)h(x)]' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)
 
Steven G said:
Can you please explain that as I'm not understanding it.
Click to expand...
P3(x) has this form:
expression1 + expression2 + expression3 + f(x1)
Each expressionn above contain (x-a) as factor.

When x=a, that factor become zero: (a-a). Therefore, all expressionn go to zero.

So x=a make P3(a)=f(x1)
 
Bruce said:
P3(x) has this form:
expression1 + expression2 + expression3 + f(x1)
Each expressionn above contain (x-a) as factor.

When x=a, that factor become zero: (a-a). Therefore, all expressionn go to zero.

So x=a make P3(a)=f(x1)
Click to expand...
But that's irrelevant to the derivative; the values of those terms are zero for that particular value of x, but that doesn't make their derivatives zero.

Tuta said:
Hi everybody,

I need to find the formula for P'(a) but I don't know how to derive this third-degree polynomial function:

1pwFnDV.png


a and b are constants but those f(x1)...f(x4) confuse me a lot. Any ideas?
Click to expand...
I would give temporary names to the big constant expressions: [math]P_3(x)=A(x-a)(x-K)(x-L)+B(x-a)(x-K)+C(x-a)+D[/math]
Now either use the product rule, or expand and just differentiate the polynomial (which is probably more work). At the end, replace A, B, C, D, K, and L with their expressions. if necessary, and then replace x with a.
 
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