Differentiable functions

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Wolfseizu

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Hi all,
Need some help here as helping a friend and it has been years since I had calculus but think I remember this having something to do with the quotient rule?

The function g(x) is differentiable. The function h(x) is defined as: h(x)=g(x)/6(x)
g(1)=-6 and g’(1)=-10
What is h’(1)?

Any help and explanation would be awesome as I would like to be able to explain it versus just getting the answer. Thank you so much!
 
Do you know how to take the derivative using the "quotient rule?" Can you show us?

-Dan
 
Wolfseizu said:
Hi all,
Need some help here as helping a friend and it has been years since I had calculus but think I remember this having something to do with the quotient rule?

The function g(x) is differentiable. The function h(x) is defined as: h(x)=g(x)/6(x)
g(1)=-6 and g’(1)=-10
What is h’(1)?

Any help and explanation would be awesome as I would like to be able to explain it versus just getting the answer. Thank you so much!
Click to expand...
Can we assume that you understand and can do the derivative of a quotient? That is what we have if I am reading the post correctly.
If it is [imath]h(x)=\dfrac{g(x)}{6x}[/imath] then [imath]h^{\prime}(x)=\dfrac{g^{\prime}(x)[6x]-g(x)[6]}{(6x)^2}[/imath].
Can you finish?

[imath] [/imath][imath][/imath][imath][/imath][imath][/imath]
 
I would consider "g(x)/6x" very bad notation for (g(x)/6)x and conider only g(x)/(6x).

Wolfseizu, the "quotient rule" says that the derivative of g(x)/f(x) is (g'(x)f(x)- g(x)f'(x))/(f(x))^2.
Here, f(x)= 6x so f'(x)= 6. The derivative of g(x)/(6x) is (6xg'(x)- 6g(x))/(36x^2)= (xg'(x)- g(x))/6x^2.

You could also write this as (1/6)x^{-1}g(x) and use the "product rule"- the derivative of f(x)g(x) is f'(x)g(x)+ f(x)g'(x). The derivative of (1/6)x^{-1}g(x) is (1/6)(-x^{-2}g(x)+ x^{-1}g'(x))= (1/6)(-g(x)/x^2+ g'(x)/x)= (1/6)(-g(x)+ xg'(x))/x^2= (1/6)(xg'(x)- g(x))/x^2 as above.

(g/(6x))'(1)= (g'(1)- g(1))/6.
 
HallsofIvy said:
I would consider "g(x)/6x" very bad notation for (g(x)/6)x and conider only g(x)/(6x).

Wolfseizu, the "quotient rule" says that the derivative of g(x)/f(x) is (g'(x)f(x)- g(x)f'(x))/(f(x))^2.
Here, f(x)= 6x so f'(x)= 6. The derivative of g(x)/(6x) is (6xg'(x)- 6g(x))/(36x^2)= (xg'(x)- g(x))/6x^2.

You could also write this as (1/6)x^{-1}g(x) and use the "product rule"- the derivative of f(x)g(x) is f'(x)g(x)+ f(x)g'(x). The derivative of (1/6)x^{-1}g(x) is (1/6)(-x^{-2}g(x)+ x^{-1}g'(x))= (1/6)(-g(x)/x^2+ g'(x)/x)= (1/6)(-g(x)+ xg'(x))/x^2= (1/6)(xg'(x)- g(x))/x^2 as above.

(g/(6x))'(1)= (g'(1)- g(1))/6.
Click to expand...
You may consider "g(x)/6x" very bad notation for (g(x)/6)x (and I agree with you) but a calculator would disagree with us. That is why I would insist that a student puts parenthesis around the entire denominator.
 
HallsofIvy said:
I would consider "g(x)/6x" very bad notation for (g(x)/6)x and conider only g(x)/(6x).
Click to expand...
Halls, I have a great amount of respect for you but I disagree with this statement. It is necessary to be clear when asking a question. The fact that several professors I've seen (and textbooks) can't get their heads around PEDMAS or its equivalents is, to my mind, very sad. I try not to be too anal about it but I try to point out the problem whenever I see it. I believe that you are correct in your assumption but if a student needs to verify with their instructor that there is a likely typo in g(x)/6x then so be it. The instructor should have spotted this.

And I always tell my students that if they see something that can be interpreted in more than one way on an exam they need to ask about it.

-Dan
 
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