Differentiating f(r): Solving a Problem

[EngnrMatt]

The problem I am struggling with is differentiating f(r)=r/sqrt(r^2 + 1)

I rewrote it as: r(r^2 +1)^(-1/2)
Split this up to get:
g(r)=r
h(r)=(r^2 + 1)^(-1/2)

By product rule, f'(r)=gh'+hg'

Obviously
g' = 1

Now by chain rule, I find h'= -r(r^2 + 1)^-(3/2)

Finally, I get f'= [(r^2 + 1)^(-1/2)] - [(r^2)((r^2 + 1)^(-3/2))]

However, Several sources say this is not correct. I followed what I believe to be the right procedures, so any feedback on where I went wrong would be great. I'm doing this from mobile, so I can't show quite as much of my work as I want. Thanks in advance.

 

[micromass]

The problem I am struggling with is differentiatingf(x)=r/sqrt(r^2 + 1)

I rewrote it as: r(r^2 +1)^(-1/2)
Split this up to get:
g(x)=r
h(x)=(r^2 + 1)^(-1/2)

By product rule, f'(x)=gh'+hg'

Obviously
g' = 2r

This isn't true.

Now by chain rule, I find h'= -r(r^2 + 1)^-(3/2)

This is true.

Finally, I get f'= [(r^2 + 1)^(-1/2)] - [(r^2)((r^2 + 1)^(-3/2))

This isn't true (likely because you have the incorrect form of g.

 

[micromass]

The problem I am struggling with is differentiatingf(x)=r/sqrt(r^2 + 1)

Also, did you mean to write $f(r)=...$ or $f(x)=...$.
If you write

$$f(x)=\frac{r}{\sqrt{r^2+1}}$$

then r is a constant, so f'(x)=0.

 

[EngnrMatt]

I meant g' = 1, about to edit it. Everything else was in fact calculated assuming g'=1, it was simply a typing error

 

[EngnrMatt]

Also, did you mean to write $f(r)=...$ or $f(x)=...$.
If you write

$$f(x)=\frac{r}{\sqrt{r^2+1}}$$

then r is a constant, so f'(x)=0.

fixed this too. it's f(r)

 

[micromass]

fixed this too. it's f(r)

OK. But please don't edit your posts have there have been replies. It makes the thread difficult to read for future readers. Just make a new post with the corrections.

Anyway:

The problem I am struggling with is differentiating f(r)=r/sqrt(r^2 + 1)

I rewrote it as: r(r^2 +1)^(-1/2)
Split this up to get:
g(r)=r
h(r)=(r^2 + 1)^(-1/2)

By product rule, f'(r)=gh'+hg'

Obviously
g' = 1

Now by chain rule, I find h'= -r(r^2 + 1)^-(3/2)

Finally, I get f'= [(r^2 + 1)^(-1/2)] - [(r^2)((r^2 + 1)^(-3/2))]

OK. Looking back at this, I think that it was correct after all. What sources say that it is incorrect??

 

[EngnrMatt]

Solutions in the back of the book, wolframalpha. Apparently the answer is (r^2 +1)^(-3/2)

 

[micromass]

Solutions in the back of the book, wolframalpha. Apparently the answer is (r^2 +1)^(-3/2)

Your answer is the same as that answer. Try to add up the fractions in

$$\frac{1}{\sqrt{r^2 + 1}} - \frac{r^2}{\sqrt{(r^2+1)^3}}$$

 

[EngnrMatt]

Thanks! I didn't think to do that, thank you so much!