Differentiate

[Isabella]

Here is a problem that I am stuck on as well. Can anyone please help me get started?

Differentiate: y= ((sqrt of x) + 5) / ((sqrt of x) - 5))

 

[Deleted member 4993]

Here is a problem that I am stuck on as well. Can anyone please help me get started?

Differentiate: y= ((sqrt of x) + 5) / ((sqrt of x) - 5))

\(\displaystyle h(x) \ = \ \dfrac{f(x)}{g(x)} \)

\(\displaystyle h'(x) \ = \ \dfrac{f'(x) * g(x) \ - \ f(x) * g'(x)}{g^2(x)} \)

Exactly where are you stuck? ...

 

[pka]

Here is a problem that I am stuck on as well. Can anyone please help me get started?
Differentiate: y= ((sqrt of x) + 5) / ((sqrt of x) - 5))

First, know the quotient rule: \(\displaystyle \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{u'v - uv'}}{{{v^2}}}\)

If \(\displaystyle y = \sqrt u + c\) then \(\displaystyle y' = \dfrac{1}{2\sqrt u}\)

Now you show us how to put it all together.

 

[simamura]

Here is a problem that I am stuck on as well. Can anyone please help me get started?

Differentiate: y= ((sqrt of x) + 5) / ((sqrt of x) - 5))

You need to use quotient rule: \(\displaystyle \displaystyle\left( \frac{\sqrt{x}+5}{\sqrt{x}-5} \right)' = \frac{(\sqrt{x}+5)'(\sqrt{x}-5)-(\sqrt{x}+5)(\sqrt{x}-5)'}{(\sqrt{x}-5)^2}=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}-5)-(\sqrt{x}+5)\frac{1}{2\sqrt{x}}}{(\sqrt{x}-5)^2}=-\frac{5}{\sqrt{x}(\sqrt{x}-5)^2}\)

 

[Isabella]

First, know the quotient rule: \(\displaystyle \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{u'v - uv'}}{{{v^2}}}\)

If \(\displaystyle y = \sqrt u + c\) then \(\displaystyle y' = \dfrac{1}{2\sqrt u}\)

Now you show us how to put it all together.

I appreciate your help pka and I would love to be able to pull it all together, I just don't know how. I am taking an online class and the response from my teacher is not too quick. If I could have help writing the first step down, I believe I could finish it.

 

[pka]

I am taking an online class and the response from my teacher is not too quick. If I could have help writing the first step down, I believe I could finish it.

Well now I understand why you were confused in another thread.
I think it is a bad idea to take an online calculus course. I have known many students who had taken AP calculus but did not do well enough to get college credit. They chose to take calculus online and did quite well. But on the other hand, students who had a weaker background had a horrible experience with online calculus.

 

[lookagain]

Here is a problem that I am stuck on as well. > > > Can anyone please help me get started? < <<

Differentiate: y= ((sqrt of x) + 5) / ((sqrt of x) - 5))

You need to use quotient rule: \(\displaystyle \displaystyle\left( \frac{\sqrt{x}+5}{\sqrt{x}-5} \right)' = \frac{(\sqrt{x}+5)'(\sqrt{x}-5)-(\sqrt{x}+5)(\sqrt{x}-5)'}{(\sqrt{x}-5)^2}=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}-5)-(\sqrt{x}+5)\frac{1}{2\sqrt{x}}}{(\sqrt{x}-5)^2}=-\frac{5}{\sqrt{x}(\sqrt{x}-5)^2}\)

simamura,

the OP was given a chance to use the help from the post before yours and post.
You don't want to post a full solution at this point as you did before the OP had a chance
to respond, especially as the OP asked to get "started." I'm speaking of myself being in
the same category as you as someone trying to help the student.

 

[Isabella]

simamura,

the OP was given a chance to use the help from the post before yours and post.
You don't want to post a full solution at this point as you did before the OP had a chance
to respond, especially as the OP asked to get "started." I'm speaking of myself being in
the same category as you as someone trying to help the student.

I think I figured this one out. Thanks so much for the help with getting started:
Answer,
dy/dx = -5 (sqrt x) / (sqrt x -5)^2

 

[Quaid]

I think I figured this one out.

dy/dx = -5 (sqrt x) / (sqrt x -5)^2

I'm thinking that the factor shown in red ought to be in the denominator.

Do you agree?

 

[Isabella]

I'm thinking that the factor shown in red ought to be in the denominator.

Do you agree?

Ok- so it should read: -5/((sqrt of x)(sqrt of x)-5^2)

Wow, not sure what I did?

Thank you so much for your help.