Deriving Fractions: Simplification Rules Explained

[ghostbuster25]

Does anyone know how to derive a fraction?

I want to be able to differentiate f(x)=$$\frac{x(20-x)}{36}$$

Ill show you what I've done but think I've just simplified it.

f(x)=$$\frac{x(20-x)}{36}$$

f(x)=$$\frac{(20-x-x)}{36}$$

f(x)=$$\frac{(10-x)}{18}$$

Is this correct?...although not sure what the rule I've followed is :(

 

[Dickfore]

Actually, it's not a fraction, since the argument occurs only in the numerator. Therefore, it is simply a polynomial with coefficients
that are fractions.

BTW, your answer is correct.

 

[LCKurtz]

Does anyone know how to derive a fraction?

I want to be able to differentiate f(x)=$$\frac{x(20-x)}{36}$$

Ill show you what I've done but think I've just simplified it.

f(x)=$$\frac{x(20-x)}{36}$$

f(x)=$$\frac{(20-x-x)}{36}$$

Nope. It is equal to

$$\frac{20x -x^2}{36}$$

f(x)=$$\frac{(10-x)}{18}$$

Is this correct?...although not sure what the rule I've followed is :(

That last expression is the correct answer for the derivative, but only the Shadow knows why it all worked out to be the derivative, and why aren't you calling it f'(x)?

 

[ghostbuster25]

Im not sure what i have done then. I understand what you mean and how to differntiate polynominals(not very well) but i think what is throwing me is no exponent like x^3=3x^2 as its deivative.
Is my solution correct?

 

[Dickfore]

That last expression is the correct answer for the derivative, but only the Shadow knows why it all worked out to be the derivative.

Actually, he used the product rule for $x^{2} = x \cdot x$.

 

[LCKurtz]

Actually, he used the product rule for $x^{2} = x \cdot x$.

Yeah, I see now what he did. And he didn't call it f'(x) when he took the derivative.

 

[ghostbuster25]

I get it!

ok let me show you to mke sure

f(x)=$$\frac{x(20-x)}{36}$$

f(x)=$$\frac{(20x-x^2)}{36}$$

f(x)=$$\frac{(20x-2x)}{36}$$

f'(x)=$$\frac{1x}{2}$$

that correct? although not f(x)=$$\frac{(10-x)}{18}$$ as you said...??

 

[LCKurtz]

I get it!

ok let me show you to mke sure

f(x)=$$\frac{x(20-x)}{36}$$

f(x)=$$\frac{(20x-x^2)}{36}$$

f(x)=$$\frac{(20x-2x)}{36}$$

That step isn't correct. If you are differentiating you should call it f'(x). And the derivative of 20x isn't 20x.

 

[ghostbuster25]

ok so derivative of 20x is 20...derivative of x^2 is 2x?
would that make it 20-2x/36...which is 20-2x/36 which is 1/-2x/18

that correct? sorry i know I am going the long way about this i just really want to understand what I am doing :)

 

[Char. Limit]

ok so derivative of 20x is 20...derivative of x^2 is 2x?
would that make it 20-2x/36...which is 20-2x/36 which is 1/-2x/18

that correct? sorry i know I am going the long way about this i just really want to understand what I am doing :)

If by "1/-2x/18" you meant $$\frac{10-2x}{18}$$, then yes.

 

[Dickfore]

Also, please note that 'derive' is not the correct term. In English, to derive means deduce a formula by doing some algebraic manipulations. The process of finding the derivative is called differentiation.

 

[ghostbuster25]

ha ha yes i did mean that. Thankyou for your help. Did LCKurtz make a mistake when he said 10-x/36 was correct?

 

[LCKurtz]

ha ha yes i did mean that. Thankyou for your help. Did LCKurtz make a mistake when he said 10-x/36 was correct?

No, I didn't. (20-2x)/36 = 2(10-x)/36 = (10-x)/18.

 

[Mu naught]

Before you even differentiate, pull the 1/36 out as a constant, and distribute the X to get:

20X - X²

then differentiate that, and multiply by the constant again and simplify.

 

[ghostbuster25]

ok thanks, sorry for ignorance but...why did you do that and not leave it as 2x?

 

[Mu naught]

Because X(20-X) = 20X - X²