Evaluate L'(9) Given L(x)=sqrt(x)*f(x), f(9)=3, F'(9)=(-4)

[UrbanXrisis]

if
L(x)=sqrt(x)*f(x)
f(9)=3
F'(9)=(-4)

find L'(9)

I think I need to use the product rule:

L'(9)=0.5x^(-0.5)*f(x)+f'(x)*sqrt(x)
L'(9)=(1/6)*3+(-4)*3
L'(9)=-11.5

Also, what if the question asked for f^3(x), how would I set that up?

 

[maverick280857]

More generally if I let f(x) and g(x) denote two functions then the first derivative of their product, viz. g(x)f(x) is written as

$$\frac{d[gf]}{dx} = g(x)\frac{df}{dx} + f(x)\frac{dg}{dx}$$

you seem to have applied the product rule correctly. I believe you mean f'(3) or f(x)^3. Is it so?

You know that f(x = 9) = 3 and f'(x = 9) = -4. Do you think this information is sufficient to fix f'(x = 3) or f(x = 3)?

 

[wais]

To find f^3(x), you would need to use the chain rule. The general formula for the nth derivative of a function f(x) is:

f^n(x) = (d^n/dx^n)(f(g(x))) * (d/dx)(g(x))

Where g(x) is the inner function within f(x). In this case, f(x) = sqrt(x)*f(x) and g(x) = x. So, to find f^3(x), we would do the following:

f^3(x) = (d^3/dx^3)(sqrt(x)*f(x)) * (d/dx)(x)
f^3(x) = (1/8x^(-3/2))*3 * 1
f^3(x) = 3/8x^(-3/2)