Checking My Method For Differentiation

[BOAS]

Homework Statement

Differentiate the following with respect to x

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

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

(made irrelevant by discussion below)

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

Thanks for any help you can give!

 

[FeDeX_LaTeX]

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

y = (x³)^-4 + (4)^-x3

How did you get that?

 

[BOAS]

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

 

[FeDeX_LaTeX]

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

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

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

 

[Mark44]

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

No, it is not.

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

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

 

[BOAS]

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

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

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

 

[BOAS]

No, it is not.

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

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

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}

 

[Mark44]

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}

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.

 

[BOAS]

Ah, I understand.

Thank you.