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

[Tuta]

Hi everybody,

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

a and b are constants but those f(x1)...f(x4) confuse me a lot. Any ideas?

 

[Bruce]

When x=a, the three factors (x-a) above are zero. That make P₃(a)=f(x₁) and x₁ is constant.

 

[Steven G]

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]

When x=a, the three factors (x-a) above are zero. That make P₃(a)=f(x₁) and x₁ is constant.

Can you please explain that as I'm not understanding it.

 

[Bruce]

Can you please explain that as I'm not understanding it.

P₃(x) has this form:
expression₁ + expression₂ + expression₃ + f(x₁)
Each expression_n above contain (x-a) as factor.

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

So x=a make P₃(a)=f(x₁)

 

[Dr.Peterson]

P₃(x) has this form:
expression₁ + expression₂ + expression₃ + f(x₁)
Each expression_n above contain (x-a) as factor.

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

So x=a make P₃(a)=f(x₁)

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.

Hi everybody,

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

a and b are constants but those f(x1)...f(x4) confuse me a lot. Any ideas?

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.