Calculus Question

[Haberdasher]

Ok, here's the problem.

Let f and g and their inverses f^(-1) and g^(-1) have the following values at x=1, x=2.

At x=1,
f(x)=3
g(x)=2
f'(x)=5
g'(x)=4

At x=2,
f(x)=2
g(x)=pi
f'(x)=6
g'(x)=7

Ok. The questions are to find the derivative of f+g at x=2, f times g at x=2, and so on. Where do I start? I really just need a hint...and fast!

 

[jman19]

Originally posted by: Haberdasher
Ok, here's the problem.

Let f and g and their inverses f^(-1) and g^(-1) have the following values at x=1, x=2.

At x=1,
f(x)=3
g(x)=2
f'(x)=5
g'(x)=4

At x=2,
f(x)=2
g(x)=pi
f'(x)=6
g'(x)=7

Ok. The questions are to find the derivative of f+g at x=2, f times g at x=2, and so on. Where do I start? I really just need a hint...and fast!

n/m even better, you already have the slopes

Do you remember the derivative rules for addition, mult, etc?

 

[Haberdasher]

I can look them up...this is a take home quiz, but it is open book and open note. Please elaborate for once I have found them...I assumed all you could do was take the slopes.

 

[Goosemaster]

derivative = instantaneous slope.

That should be enough;

 

[Goosemaster]

Originally posted by: Haberdasher
I can look them up...this is a take home quiz, but it is open book and open note. Please elaborate for once I have found them...I assumed all you could do was take the slopes.

You kind of shut everyone out of the disscussion by saying that it is a take home quiz.

I'm sorry, but if you failed to study, you should take the failure like a man or read your damn book.


TRUST ME. I failed Calc 4 times before I understood that I wasn't going anywhere without studying.

Once I realized that I was fine.

 

[jman19]

Originally posted by: Haberdasher
I can look them up...this is a take home quiz, but it is open book and open note. Please elaborate for once I have found them...I assumed all you could do was take the slopes.

This should get you started:

(f+g)' = f'+g'

(fg)' = fg' + f'g

 

[Haberdasher]

I didn't think that the derivative could be obtained from 2 points, just because they are sequential. That's the average slope. The derivative is merely the best linear approximation of the curve at a point. Just because I can get a linear equation from 2 points close to each other doesn't necessarily mean I have the derivative...

Right?

 

[Haberdasher]

Originally posted by: Goosemaster

Originally posted by: Haberdasher
I can look them up...this is a take home quiz, but it is open book and open note. Please elaborate for once I have found them...I assumed all you could do was take the slopes.

You kind of shut everyone out of the disscussion by saying that it is a take home quiz.

I'm sorry, but if you failed to study, you should take the failure like a man or read your damn book.


TRUST ME. I failed Calc 4 times before I understood that I wasn't going anywhere without studying.

Once I realized that I was fine.

I'm a good student in this class. I have an A. I just spent about an hour trying to find a solution for this problem using the knowledge I have, my notes, my book, etc. I'm not lazy. I'm working my butt off for one problem.

 

[Howard]

Originally posted by: jman19

Originally posted by: Haberdasher
I can look them up...this is a take home quiz, but it is open book and open note. Please elaborate for once I have found them...I assumed all you could do was take the slopes.

This should get you started:

(f+g)' = f'+g'

(fg)' = fg' + f'g

 

[Haberdasher]

Thanks for the assistance, I'm looking up all the rules for the derivatives when you add, subtract, multiply, etc...

Not done, but I think I'm out of the woods. I swear, my brain just wasn't functioning after hours of math worksheets.

 

[chuckywang]

Originally posted by: Haberdasher
I didn't think that the derivative could be obtained from 2 points, just because they are sequential. That's the average slope. The derivative is merely the best linear approximation of the curve at a point. Just because I can get a linear equation from 2 points close to each other doesn't necessarily mean I have the derivative...

Right?

You can't obtain the derivatives from just two points. However, in your problem, you already have the derivatives at two points.

 

[Haberdasher]

Originally posted by: chuckywang

Originally posted by: Haberdasher
I didn't think that the derivative could be obtained from 2 points, just because they are sequential. That's the average slope. The derivative is merely the best linear approximation of the curve at a point. Just because I can get a linear equation from 2 points close to each other doesn't necessarily mean I have the derivative...

Right?

You can't obtain the derivatives from just two points. However, in your problem, you already have the derivatives at two points.

Ok, cool. I think my problem is one of someone who can get concepts if he studies them, but can't recognize them easily elsewhere (not a good type for math, eh?). We've never been given a type of problem like this before (using already solved functions,) and having to use the chain rule, quotient rule, etc. on 2 functions that don't have variables in them threw me off.

A hearty thank-you, again.