1. Chain Rule!!

1. Given y=f(x) with f(1)=4 and f'(1)=3, find

a) g'(1) if g(x)=(f(x))^(1/2)

b) h'(1) if h(x)= f ((x)^(1/2))

For part a,

g'(1)=1/2 (f(x))^-1/2 (f'(x))
=(1/2)(4^-1/2)(3)=.75

Is this correct?

For part b,

h'(x)=f(1/2)(x^-1/2)

I am unsure if part b's setup is even right...if it is what do I plug in for f? Please help me out soon!

2. Let f(v) be the gas consumption (in liters/km) of a car going at velocity v (i km/r). In other words, f(v) tells you how many liters of gas the car uses to go one km, if it is going at velociy v. You are told that...

f(80)=.05 and f'(80)=.0005

a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) and g(v)? Find g(80) and g'(80)

b) Let h(v) be te gas consumption in liters per hour. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the relationship beween h(v) and f(v)? Find h(80) and h'(80).

c) How would you explain the practical meaning of the values of these functions and their derivatives to a driver who knows no calculus?

I have no idea where to start and I don't exactly understand the questions. Help/guidance/hints would be appreciated greatly! Thanks in advance

2. Originally Posted by bluea32
1. Given y=f(x) with f(1)=4 and f'(1)=3, find

a) g'(1) if g(x)=(f(x))^(1/2)

b) h'(1) if h(x)= f ((x)^(1/2))

For part a,

g'(1)=1/2 (f(x))^-1/2 (f'(x))
=(1/2)(4^-1/2)(3)=.75

Is this correct? Mr F says: Yes.

For part b,

h'(x)=f(1/2)(x^-1/2)

Mr F says: Using the chain rule, $h'(x) = f'\left( x^{1/2} \right) \left( \frac{1}{2} x^{-1/2} \right) = f'\left( \sqrt{x} \right) \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$.
So $\, h'(1) = f'\left( \sqrt{1} \right) \frac{1}{2 \sqrt{1}} = f'(1) \frac{1}{2} = \frac{3}{2}$.

I am unsure if part b's setup is even right...if it is what do I plug in for f? Please help me out soon!

2. Let f(v) be the gas consumption (in liters/km) of a car going at velocity v (i km/r). In other words, f(v) tells you how many liters of gas the car uses to go one km, if it is going at velociy v. You are told that...

f(80)=.05 and f'(80)=.0005

a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) and g(v)? Find g(80) and g'(80)

b) Let h(v) be te gas consumption in liters per hour. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the relationship beween h(v) and f(v)? Find h(80) and h'(80).

c) How would you explain the practical meaning of the values of these functions and their derivatives to a driver who knows no calculus?

I have no idea where to start and I don't exactly understand the questions. Help/guidance/hints would be appreciated greatly! Thanks in advance
Q2 a) f(v) is in units of liter/km. g(v) is in units of km/liter. Inspection of the units suggests that $g(v) = \frac{1}{f(v)} = [f(v)]^{-1}$ ....

To get g'(v), you could use the chain rule: $g'(v) = -[f(v)]^{-2} \cdot f'(v) = -\frac{f'(v)}{[f(v)]^2}$ ....

b) h(v) is in units of litre/hour. f(v) is in units of liter/km. Therefore $\frac{h(v)}{f(v)}$ is in units of km/hr. Since km/hr is the unit of velocity, this suggests that $\frac{h(v)}{f(v)} = v \Rightarrow h(v) = v \cdot f(v)$.

To get h'(v), you'd use the product rule: $h'(v) = f(v) + v \cdot f'(v)$ .......

c) I leave you to think about ......

3. hey thank you SO much for the help! I just have a few more questions...

1) What is the label/unit for g'(v)

2) What is the label/unit for h'(v)

Thanks again =)

4. Originally Posted by bluea32
hey thank you SO much for the help! I just have a few more questions...

1) What is the label/unit for g'(v)

2) What is the label/unit for h'(v)

Thanks again =)
Well, g'(v) is the rate of change of g with resepct to v. You know the unit for g and you know the unit for v. So ...... divide the units ......

Ditto h'(v).