Question regarding a derivative

[in the rye]

Homework Statement

Write the derivative of y = (x²+4x+3)/(x^1/2)

I got the correct answer, but my question is, why can't I rewrite this as:

y = (x^2+4x+3)*(1/x^1/2)

Then see my attempted solution for the result...

Homework Equations

y = (x²+4x+3)/(x^1/2)

The Attempt at a Solution

y = (x^2+4x+3)*(x^-1/2)
d/dx = (2x+4)*(-1/2x^-3/2)The problem is that this seems to drop out a term from the equation, because the answer that I got when treating each term as being divided by x^1/2 gave me:

d/dx = 1.5x^.5+2x^-.5-3/(2x(x^.5)), which was the answer in the book. Yeesh, sorry this is difficult to read over forums. I tried to use decimals to make it more readable over the forums, but I don't know if it helped. Maybe next time I'll post an image of my work. Sorry.

Anyways, when I did the method I show above, I get a different equation entirely because the term drops out due to the constant becoming a 0 in the derivative. However, it seems that I've followed all the laws correctly that I've been taught. That is, why can't I take each term as a derivative, and multiply it by another derivative?

 

[axmls]

Because then you need to use the product rule. If you do that, you'll get the right answer.

 

[in the rye]

Okay, thanks. I just realized that I am going to come across that in the next section. :D

 

[Mark44]

y = (x^2+4x+3)*(x^-1/2)
d/dx = (2x+4)*(-1/2x^-3/2)

Two comments:

1. In your second equation what you have is the product of the derivatives of the factors in the first equation. That's not how the product rule works. IOW, $\frac d {dx} f(x) \cdot g(x) \ne f'(x) \cdot g'(x)$
2. In your second equation you have "d/dx = ..." This is incorrect. The symbol $\frac d {dx}$ is an operator that requires something to operate on, and that appears just to the right of this operator. For the second equation, "dy/dx = " would be appropriate, but "d/dx = " is not, since you haven't shown what it is that you're taking the derivative of. It's a bit like writing √ = 3 or sin = 0.5.

 

[Ray Vickson]

Homework Statement

Write the derivative of y = (x²+4x+3)/(x^1/2)

I got the correct answer, but my question is, why can't I rewrite this as:

y = (x^2+4x+3)*(1/x^1/2)

Then see my attempted solution for the result...

Homework Equations

y = (x²+4x+3)/(x^1/2)

The Attempt at a Solution

y = (x^2+4x+3)*(x^-1/2)
d/dx = (2x+4)*(-1/2x^-3/2)The problem is that this seems to drop out a term from the equation, because the answer that I got when treating each term as being divided by x^1/2 gave me:

d/dx = 1.5x^.5+2x^-.5-3/(2x(x^.5)), which was the answer in the book. Yeesh, sorry this is difficult to read over forums. I tried to use decimals to make it more readable over the forums, but I don't know if it helped. Maybe next time I'll post an image of my work. Sorry.

Anyways, when I did the method I show above, I get a different equation entirely because the term drops out due to the constant becoming a 0 in the derivative. However, it seems that I've followed all the laws correctly that I've been taught. That is, why can't I take each term as a derivative, and multiply it by another derivative?

Please, please do NOT post an image; a typed version is much better, and is the PF standard.

Anyway, it might be easiest of all if you re-write your function before doing the derivative:
y = \frac{x^2 + 4x + 3}{x^{1/2}} = \frac{x^2}{x^{1/2}} + \frac{4x}{x^{1/2}} + \frac{3}{x^{1/2}} \\<br /> \\<br /> = x^{3/2} + 4 x^{1/2} + 3 x^{-1/2}
In this last form you can just differentiate term-by-term, and so do not need the product or quotient rules.

 

[Fredrik]

The derivative should be denoted by dy/dx, not d/dx. dy/dx means f'(x), where f is the function defined implicitly by the previously stated relationship between x and y.