# Homework Help: Checking My Method For Differentiation

1. Oct 22, 2013

### BOAS

1. The problem statement, all variables and given/known data

Differentiate the following with respect to x

y = $\frac{4}{x^{3}}$ + $\frac{x^{3}}{4}$

3. The attempt at a solution

So the problem here is really getting this into a form that is easy to differentiate and i'd just like to show what i'm doing before I go ahead and do the rest of my questions. The step i'm least confident with is applying chain rule to each term separately...

I want to use the chain rule here, so I would rearrange to this;

y = (x3)-4 + (4)-x3

dy/dx = -4(x3)-53x2 - x3(4)-x3-1.0

dy/dx = -4(x3)-53x2

dy/dx = -12x2(x3)-5

So, is this ok? I am concerned that my second term disappears, but that's what it looks like I have to do.

Last edited: Oct 22, 2013
2. Oct 22, 2013

### FeDeX_LaTeX

How did you get that?

3. Oct 22, 2013

### BOAS

$(x^{3})^{-4}$ = $4/x^{3}$ is it not?

4. Oct 22, 2013

### FeDeX_LaTeX

It is not -- $(x^3)^{-4} = x^{-12}$.

$\frac{4}{x^3} = 4x^{-3}$

5. Oct 22, 2013

### Staff: Mentor

No, it is not.

$\frac 4 {x^3} = 4x^{-3}$

For your problem there is no need for the chain rule.

6. Oct 22, 2013

### BOAS

I have made this mistake a few times before... Thanks for catching it.

7. Oct 22, 2013

### BOAS

That's become somewhat clearer to me after my mistake was pointed out.

so y = $4x^{-3} + x^{3}(4^{-1})$

dy/dx = $-12x^{-4} -4^{-2}3x^{2}$

dy/dx = $-12x^{-4} - (3/2) x^{2}$

8. Oct 22, 2013

### Staff: Mentor

y = 4x-3 + (1/4)x3
dy/dx = -12x-4 + (3/4)x2

You are using the power rule on something that is not a function. 4-1 is a constant, so its derivative is zero.

9. Oct 22, 2013

### BOAS

Ah, I understand.

Thank you.