Simplifying a differentiation problem

[fk378]

Homework Statement

[3x^2 + 2(sqrt x)] / x

My book says it simplified it to:

3x + 2x ^-1/2I understand the 3x part since 3x^2/x = 3x
But where did the second part come from? I tried to make sense with the power rule but it still doesn't seem clear to me. Anyone get it?

 

[dynamicsolo]

[[3x^2 + 2(sqrt 2)] / x

My book says it simplified it to:

3x + 2x ^-1/2

Is this supposed to be [[3x^2 + 2(sqrt x)] / x ?

Then it would make sense.

 

[fk378]

Yes you're right. Can you explain it?

 

[FocusedWolf]

There's a formula for simplying derivitives that look like big fractions like yours... i forget the name but the way it goes was: "((bottom) * derivitive[of the top] - (top)*derivitive[of the bottom])) over the bottom squared)"

like this:

(bottom * deriv[top]) - (top * deriv[bottom])
---------------------------------------------
(bottom)^2

that and square root is same as x^(1/2) power and just apply the chain rule i think it's called which makes it 1/(2*sqrt(x)).
whats important is it's different for this: (3x)^(1/2) which becomes (1/(2*sqrt(x)) * derivitive of (3x).

on side note: if you don't have a ti89, get one. it does calculus and is extremely helpful in checking calculus homework and understanding why certain things differentiate the way they do.

 

[turdferguson]

So this is at the end of the problem? Distribute the division: (a+b)/c = a/c + b/c
Remember that sqrt(x) = x^1/2

 

[fk378]

Yes I know the quotient rule and power rule, but I don't see how the first equation was simplified.

 

[fk378]

So this is at the end of the problem? Distribute the division: (a+b)/c = a/c + b/c
Remember that sqrt(x) = x^1/2

There isn't a common factor in the original equation, though...

 

[hage567]

I'm confused on what you think you are doing. Are you trying to differentiate the above expression, or just simplifying it? If you are just simplifying, you just divide the terms. No calculus involved in this step.
So for the second term you have $$\frac{2\sqrt{x}}{x}$$, which is the same as $$\frac{2{x}^\frac{1}{2}}{x}$$
Do you know your laws of exponents?

 

[fk378]

I'm confused on what you think you are doing. Are you trying to differentiate the above expression, or just simplifying it? If you are just simplifying, you just divide the terms. No calculus involved in this step.
So for the second term you have $$\frac{2\sqrt{x}}{x}$$, which is the same as $$\frac{2{x}^\frac{1}{2}}{x}$$
Do you know your laws of exponents?

The book is saying that 2(sqrt 2) / x is the same as 2x^-1/2
so while I did get the same answer as you, it apparently is not right according to the book.

 

[hage567]

The book is saying that 2(sqrt 2) / x is the same as 2x^-1/2

It is! $$\frac{2{x}^\frac{1}{2}}{x} = 2x^{\frac{1}{2}-1} = 2x^\frac{-1}{2}$$

When dividing, you subtract the exponents.

I hope I'm understanding what you're asking.

 

[fk378]

Oh I see it now...wow what a brain fart...anyways, thank you for explaining it to me clearly and in a nice manner.

 

[hage567]

lol you're welcome.