• October 20th 2008, 07:00 PM
sidhlyn
Resolved
Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

It seems like it should be simple, but I just cannot get it worked out properly!
Given: y=f(x) with f(1)=4 and f'(1)=3, find:
- (a.) g'(1) if g(x)=sq. rt. of f(x)
and then
+ (a.) g'(1) if g(x)=sq. rt. of f(x)....which works out to g'(1)=3/4
but I cannot figure out:
(b.) h'(1) if h(x)=f(sq.rt. of x)

Please, if someone could please offer me assistance, even some hints would be appreciated.
These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????

Thank you,
Lynn
• October 20th 2008, 07:08 PM
Chris L T521
Quote:

Originally Posted by sidhlyn
Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

It seems like it should be simple, but I just cannot get it worked out properly!
Given: y=f(x) with f(1)=4 and f'(1)=3, find:
(a.) g'(1) if g(x)=sq. rt. of f(x)
and then
(b.) h'(1) if h(x)=f(sq.rt. of x)

Please, if someone could please offer me assistance, even some hints would be appreciated.
These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????

Thank you,
Lynn

Note that $\frac{d}{dx}\left[\left(f(x)\right)^{\frac{1}{2}}\right]=\frac{1}{2}\left(f(x)\right)^{-\frac{1}{2}}\cdot f'(x)$

Do you think you can tackle part (a) now?

Note that $\frac{d}{dx}\left[f(u)\right]=f'(u)\cdot \frac{\,du}{\,dx}$

Do you think you can tackle part (b) now?

--Chris
• October 20th 2008, 07:39 PM
sidhlyn
Still stumped on b
Thanks Chris...
Actually I did manage to get (a) on my own after reviewing my notes again.
I was just editing the post as I was sitting here pondering over it that I got part a, & saw your note after it updated.
However, no....now I am still stumped on how to apply "u" (we use dy/du for the "inside") to part b. I'm not seeing how to apply it to f(x^1/2). ???

Thanks again,
Lynn
• October 20th 2008, 07:52 PM
Chris L T521
Quote:

Originally Posted by sidhlyn
Thanks Chris...
Actually I did manage to get (a) on my own after reviewing my notes again.
I was just editing the post as I was sitting here pondering over it that I got part a, & saw your note after it updated.
However, no....now I am still stumped on how to apply "u" (we use dy/du for the "inside") to part b. I'm not seeing how to apply it to f(x^1/2). ???

Thanks again,
Lynn

Let $u=x^{\frac{1}{2}}$

Thus, substituting this all into the formula I gave you, we see that $\frac{d}{\,dx}\left[f(u)\right]=f'(u)\cdot\frac{\,du}{\,dx}\implies\frac{d}{\,dx} \left[f(x^{\frac{1}{2}})\right]=f'(x^{\frac{1}{2}})\cdot\frac{\,d}{\,dx}\left[x^{\frac{1}{2}}\right]=f'(x^{\frac{1}{2}})\cdot\frac{1}{2\cdot x^{\frac{1}{2}}}$

Since $h'(x)=f'(x^{\frac{1}{2}})\cdot\frac{1}{2\cdot x^{\frac{1}{2}}}$, what is $h'(1)$??

--Chris
• October 20th 2008, 08:10 PM
sidhlyn
Thank you, Chris!
Oh?
I didn't think I could do that, since the exponent affects the input itself.
I'll give it a try.

Thanks again!!!

~Lynn
• December 8th 2008, 02:38 PM
mr fantastic
Quote:

Originally Posted by sidhlyn
Hello, we have just begun using the chain rule in Calculus 1, but I am stumped on a problem which is blinding me to everything else.

It seems like it should be simple, but I just cannot get it worked out properly!
Given: y=f(x) with f(1)=4 and f'(1)=3, find:
- (a.) g'(1) if g(x)=sq. rt. of f(x)
and then
+ (a.) g'(1) if g(x)=sq. rt. of f(x)....which works out to g'(1)=3/4
but I cannot figure out:
(b.) h'(1) if h(x)=f(sq.rt. of x)

Please, if someone could please offer me assistance, even some hints would be appreciated.
These are driving me insane with frustration and I have a 105% average in my class!!! What am I overlooking????

Thank you,
Lynn

Thread closed due to OP deleting original question.