Calculus Derivatives: Find (a-d) at 3

[hsd]

Homework Statement

F(3)=−2, g(3)=9, f′(3)=−2, and g′(3)=2, find the following numbers:

(a) (f+g)′(3)

(b) (fg)′(3)

(c) (f/g)′(3)

(d) (f/(f−g))′(3)

The Attempt at a Solution

I already have (a) and (b) [a=0 and b=-22]

for (c) i tried:

(g(x)*f'(x) - f(x)g'(x)) / (g(x))^2

evaluate at 3:
(9)(-2) - (-2)(2) / 4
-18 + 4 / 4
-14/4
-7/2

WHICH IS THE WRONG ANSWER

(d) I just can't get started.

 

[berkeman]

Homework Statement

F(3)=−2, g(3)=9, f′(3)=−2, and g′(3)=2, find the following numbers:

(a) (f+g)′(3)

(b) (fg)′(3)

(c) (f/g)′(3)

(d) (f/(f−g))′(3)

The Attempt at a Solution

I already have (a) and (b) [a=0 and b=-22]

for (c) i tried:

(g(x)*f'(x) - f(x)g'(x)) / (g(x))^2

evaluate at 3:
(9)(-2) - (-2)(2) / 4
-18 + 4 / 4
-14/4
-7/2

WHICH IS THE WRONG ANSWER

(d) I just can't get started.

The denominator for c should be g^2 and you plugged in g'^2 (simple error).

 

[berkeman]

And for d), can you still just use the quotient rule? (I'm not sure, but try it...)

 

[Mark44]

And for d), can you still just use the quotient rule? (I'm not sure, but try it...)

Yes, use the quotient rule. After you get the derivative, evaluate the derivative at x = 3.

 

[hsd]

Yes, use the quotient rule. After you get the derivative, evaluate the derivative at x = 3.

I think I am applying the rule wrong. This is what I did:

(f-g)'(f)-(f)'(f-g)/(f-g)
[(-2)(-2)]'[-2]-[-2]'[(-2)(-9)]/[(-2)(-9)]
8-22/-11
-14/-11 (wrong answer)

Can you please tell me what it is that I am doing wrong?

 

[Mark44]

I think I am applying the rule wrong. This is what I did:

(f-g)'(f)-(f)'(f-g)/(f-g)

You have three mistakes above:
1. The terms in the numerator are switched, which will give you the wrong sign for your answer.
2. The term in the denominator needs to be squared.
3. You are missing a pair of parentheses in the numerator.

[(-2)(-2)]'[-2]-[-2]'[(-2)(-9)]/[(-2)(-9)]
8-22/-11
-14/-11 (wrong answer)

Can you please tell me what it is that I am doing wrong?

It would be helpful if you included the arguments for the general derivative.
Let h(x) = (f/g)(x), then h'(x) = (f/g)'(x) = [g(x)f'(x) - f(x)g'(x)]/g²(x)

And then substitute 3 for x, similar to what I have done below.
So h'(a) = [g(a)f'(a) - f(a)g'(a)]/g²(a)