Simplifying a Function Prior to Finding a Derivative

[Drakkith]

Okay guys, this is driving me absolutely nuts.
I'm working on finding derivatives using the product and quotient rules and the book will sometimes simplify the problem before finding the derivative but sometimes wont and I don't understand why.

For example: The function y = (v³-2v√v)/v
The book simplifies it to v²-2v^½ and then gets 2v-v^-½ as the derivative without even using the product or quotient rules.

Then on this problem: f(x) = x/[x+(c/x)] it doesn't simplify first. I started off by simplifying it as follows:
Original function: f(x) = x/[x+(c/x)]
Making the bottom all one fraction: x/[(x²+c)/x]
Bottom and top cancel out: 1/(x²+c).
But using the quotient rule on this gives me an answer of 2x/(x²+c)², which is wrong. The answer is actually 2cx/(x²+c)², which the book gets by using the quotient rule without simplifying first.

When is it okay to simplify before finding the derivative and when isn't it?

 

[mfb]

Bottom and top cancel out

They don't. You divide the numerator by x but multiply the denominator with x.

I'm working on finding derivatives using the product and quotient rules and the book will sometimes simplify the problem before finding the derivative but sometimes wont and I don't understand why.

It probably chooses the most convenient option. A proper simplification does not change the function, you can always do it if you like.

 

[Drakkith]

Dang it, I didn't switch the fraction around when I was doing the multiplication. Sigh... it's been like this all day today...

So x/[(x²+c)/x] is equal to x²/(x²+c)
So the derivative is: (x²+c)(2x) - (x²)(2x) / (x²+c)²
Which then becomes: 2x³+2cx-2x³ / (x²+c)²
Or: 2cx/(x²+c)², which is the correct answer...

Thanks Mfb.