Differentiation help

freeze4real

hello can can you help me with this leaving cert differentiation.

hypothetically here its 1/8Csquared/C

I am thinking that I should bring the C from the bottom up so it will be 1/8C squared - C so the answer is 1/8C.

Am I close to being right ?

MathsManiac

I can't understand what you're asking.

Can you give us the exact text of the question?

freeze4real

there's no question to it.

Its just that i have to differentiate 1/8Csquared divided by C

MathsManiac

Do you mean (1 / (8 * c^2)) / c?

If so, that's the same as 1 / (8 * c^3).

What are you differentiating it with respect to?

If it's with respect to c, then you can re-write it as (1/8) * c^(-3), and then differentiate it to get (-3/8) * c^(-4),
which can be written as (-3)/(8*(c^4)).

If, on the other hand, you're differentiating it with respect to something else, and if c is constant, then it's just 0.

freeze4real

i'm differentiating with respects to C only.

**Timbuk2**

Does the question look like this?

[latex]\displaystyle \frac{\frac{1}{8c^2}}{c}[/latex]?

If so, as MathsManiac said, that can be rewritten as

[latex]\displaystyle \frac{1}{8c^3} = \frac{1}{8}c^{-3}[/latex]

Differentiate with respect to c

[latex]\displaystyle -\frac{3}{8}c^{-4}[/latex]

Which can be written as (if you want):
[latex]\displaystyle -\frac{3}{8c^4}[/latex]

Which is the same as what MathsManiac said - sometimes it's a bit easier to see in Latex notation! Are we correct about the form of the question?

freeze4real

**Timbuk2** wrote: »

Does the question look like this?

[latex]\displaystyle \frac{\frac{1}{8c^2}}{c}[/latex]?

If so, as MathsManiac said, that can be rewritten as

[latex]\displaystyle \frac{1}{8c^3} = \frac{1}{8}c^{-3}[/latex]

Differentiate with respect to c

[latex]\displaystyle -\frac{3}{8}c^{-4}[/latex]

Which can be written as (if you want):
[latex]\displaystyle -\frac{3}{8c^4}[/latex]





Which is the same as what MathsManiac said - sometimes it's a bit easier to see in Latex notation! Are we correct about the form of the question?

How did you get the 3/8c4. I am only meant to differentiate once in respect to C

**Timbuk2**

freeze4real wrote: »

How did you get the 3/8c4. I am only meant to differentiate once in respect to C

I'll probably make a mess of explaining this, but to differentiate it, crudely put in words, you basically take the exponent down and multiply, then reduce the exponent by one.

For example, the derivative of (with respect to x) [latex]x^3[/latex] is [latex]3x^2[/latex]

In general, the derivative of (with respect to x) [latex]x^n[/latex] is [latex]nx^{n-1}[/latex]

So in the above example, you want to derive [latex]\displaystyle \frac{1}{8}c^{-3}[/latex]
which is:
[latex](-3)\frac{1}{8}c^{-3-1} = \displaystyle \frac{-3}{8}c^{-4}[/latex]

Also note, the usual way of writing derivatives would be
[latex]\displaystyle{\frac{d}{dc}( \frac{1}{8}c^{-3})}[/latex]

Or if [latex]f(c) = \displaystyle \frac{1}{8}c^{-3}[/latex]
The derivative is written as [latex]f'(c)[/latex]
(" f prime of c " ).

LeixlipRed

I think he's confused about the change in the sign of the power.

freeze4real

Thanks for the help I get it now.

**Timbuk2** wrote: »

I'll probably make a mess of explaining this, but to differentiate it, crudely put in words, you basically take the exponent down and multiply, then reduce the exponent by one.

For example, the derivative of (with respect to x) [latex]x^3[/latex] is [latex]3x^2[/latex]

In general, the derivative of (with respect to x) [latex]x^n[/latex] is [latex]nx^{n-1}[/latex]

So in the above example, you want to derive [latex]\displaystyle \frac{1}{8}c^{-3}[/latex]
which is:
[latex](-3)\frac{1}{8}c^{-3-1} = \displaystyle \frac{-3}{8}c^{-4}[/latex]

Also note, the usual way of writing derivatives would be
[latex]\displaystyle{\frac{d}{dc}( \frac{1}{8}c^{-3})}[/latex]

Or if [latex]f(c) = \displaystyle \frac{1}{8}c^{-3}[/latex]
The derivative is written as [latex]f'(c)[/latex]
(" f prime of c " ).

**Timbuk2**

LeixlipRed wrote: »

I think he's confused about the change in the sign of the power.

Oh right!

Well it's a main property of exponents that [latex]x^{-a} = \displaystyle \frac{1}{x^a}[/latex]

In very simple English, you could say that a negative exponent is on the wrong part of the fraction. To change the exponent from a negative to a positive, simply take it to the other side of the fraction.

So I have a [latex]c^{-4}[/latex]. It might be a bit simpler to take this to the denominator of the fraction, where it will become [latex]\displaystyle \frac{1}{c^4}[/latex]. But you don't have to do this, it's no more correct and it doesn't change anything.

If you wanted to calculate a negative exponent, first calculate the positive exponent, then take the reciprocal.

For example, if you wanted to calculate [latex]2^{-3}[/latex]. This may not be immediately obvious, as intuitively, it's hard to think of raising something to the minus third power. But we know that 2^3 is 8. Therefore [latex]2^{-3}[/latex] is [latex]\frac{1}{8} = 0.125[/latex].