# Double derivative, really hard need help

1. Jun 8, 2006

### ozzie6616

Okay so I have this calculus teacher who is crazy and gave us this problem on an exam. I don't know how to figure it out and I tried getting explanations but they were really hard and complicated. So here is the problem

x^2 [f(x)]^4 + xf(x) = 6

f(2) = 1 Find f''(2)

I would most greatly appreciate it if someone showed me how to solve the problem step by step.

2. Jun 8, 2006

### neutrino

Hi ozzie,
I hope you know how to differentiate using the product rule and chain rule, because those are the only things you need to solve the problem. Try that out and report any difficulty you might have.

3. Jun 8, 2006

### arildno

I assume it is meant to differentiate the equation first one; from that determine f'(2), and then differentiate again and determine f''(2).

4. Jun 8, 2006

### jasc15

that looks like it requires some implicit differentiation, which i havent encountered since calc I, so i dont really know how to do that one.

5. Jun 8, 2006

### dimensionless

Is one method preferable to the other?

6. Jun 8, 2006

### neutrino

Both are required.

7. Jun 8, 2006

### ozzie6616

yes it does require implicit differentiation. I have gone through the first step to find the first derivative, however when I get stumped when finding the second derivative.

This is how far i got and I don't know if its right.
2x[f(x)]^4 + 4[F(x)]^3[f'(x)]x^2 + f(x) + f(x) + f'(x)x=0

Now I don't know what to do next.

8. Jun 8, 2006

### neutrino

There is an extra f(x). Now substitute x = 2 and f(2) = 1 to find f'(x). And differntiate the above equation once more to get one the expression for f''(x). Plug-in the known values at x=2 and find f''.

9. Jun 8, 2006

### lance

so, the general plan is:
-differentiate once with respect to x
-evaluate the expression @x=2
-solve for f '(2)
-differentiate again with respect to x
-evaluate the new expression @x=2
-solve for the only unknown f ''(2)

a tip:

d/dx[ f(x)^4 ] = 4*f '(x)*f(x)^3

so the first derivative should be:

x^2*4*f '(x)*f(x)^3+2x*f(x)^4+x*f '(x)+f(x)=0

that is using the product rule on both terms of the original equation.

to further understand the differentiation of a general function f(x)
take careful note that: d/dx[f(x)]=f '(x)*d/dx(x)
so for example d/dx[ f(x^2) ]= f '(x^2)*2x