Thread: [SOLVED] Power, Product, and Quotient Rules Problems

1. [SOLVED] Power, Product, and Quotient Rules Problems

I need some assistance on several problems I am absolutely stuck on:

1.
"The quantity of a drug, Q mg, present in the body t hours after the injection of the drug is given is:

Q=f(t)=80te^-0.38t

find:
f(1)= 54.71
f'(1)=______
f(6)=49.10
f'(6)=______

I know I'd find the derivative of f(t). Would it be:

80t(-.38t)e^-.38t
?

2nd question:
"A dose, D, of a drug causes a temperature change, T, in a patient. For C a positive constant, C=32, T is given by:

T=((C/2)-(D/3))D^3

What is the rate of change of temperature change with respect to dose?

dT/dD=_____

For what doses does the temperature change increase as the dose increases?

D < _____"

Not sure where to start here. Maybe substitute C=32 and take the derivative of the equation?

3rd:
(describes a museum deciding to sell a painting and invest the proceeds)
basically:
"B(t)= P(t)(1.04)^(10-t)"

{where P(t)=sale price, B(t)=balance in yr. 2010, t=year when painting is sold, measured from the year 2000 so 0<t<10}

"Find B'(7), given that P(7)=120,000 and P'(7)=5,000.

B'(7)= _____

I have tried finding the derivative of the equation {B'(t)=P(t)ln(1.04)(1.04)^(10-t)}
and evaluating for t=7, but that didn't work. Any other ideas?

Last question:
"Given r(2) = 16, s(2) = 3, s(4) = 4, r'(2) = 9, s'(2) = 2, and s'(4) = 9, compute the following derivatives, or state what additional information you would need to be able to compute the derivative.

(a.)H'(2) if H(x) = r(x) + s(x)
H'(2)= 11
(b.)H'(2) if H(x) = 8s(x)
H'(2)= 16
(c.)H'(2) if H(x) = r(x) * s(x)
H'(2)= _____
(d.)H'(2) if H(x) = sqrt(r(x))
H'(2)= _____"

Is something missing from (c) and (d) that I need? I've tried for (c):
16*3
9*2 (r'(x)*s'(x)=H'(x)?)
and (just a big guess) r(s(x))

and for (d):
256 (r(x)^2)
3 (sqrt(r'(x)))
4 (sqrt(r(x)))

but none of these worked out.

I hate to be asking so many questions in one post, but I am at a complete loss for where to go based on what I have already tried. Any help is (as always) highly appreciated!

2. Originally Posted by bpdude
I need some assistance on several problems I am absolutely stuck on:

1.
"The quantity of a drug, Q mg, present in the body t hours after the injection of the drug is given is:

Q=f(t)=80t * e^-0.38t

find:
f(1)= 54.71
f'(1)=______
f(6)=49.10
f'(6)=______

I know I'd find the derivative of f(t). Would it be:

80t(-.38t)e^-.38t <<<<<<< No. You have to use the product rule here (see above)
?

2nd question:
"A dose, D, of a drug causes a temperature change, T, in a patient. For C a positive constant, C=32, T is given by:

T=((C/2)-(D/3))D^3

What is the rate of change of temperature change with respect to dose?

dT/dD=_____

For what doses does the temperature change increase as the dose increases?

D < _____"

Not sure where to start here. Maybe substitute C=32 OK then expand the bracket and take the derivative of the equation !

...
...

3. So it would be:
80*e^-.38t?
Then substitute 1 and 6?

And the 2nd one would be:

((32/2)-(D/3))D^3
dT/dD=3(16-(D/3))D^2
dT/dD=(48-(3D/3))D^2

Would the 3D/3 simplify to "D"?:
dT/dD=(48-D)D^2

4. Originally Posted by bpdude
So it would be:
80*e^-.38t?
Then substitute 1 and 6?

And the 2nd one would be:

((32/2)-(D/3))D^3
dT/dD=3(16-(D/3))D^2
dT/dD=(48-(3D/3))D^2

Would the 3D/3 simplify to "D"?:
dT/dD=(48-D)D^2
$f'(t) = 80e^{-0.38t} - (80)(0.38)te^{-0.38t}$

5. Thank you!

Originally Posted by e^(i*pi)
$f'(t) = 80e^{-0.38t} - (80)(0.38)te^{-0.38t}$
It makes sense now (I'm not sure why I couldn't get to that)! Thank you, e^(i*pi)!

Earboth, what exactly do you mean by "expand the bracket"?

One down, 3 to go.

Any ideas about 3 and 4?

6. Originally Posted by bpdude
...

Earboth, what exactly do you mean by "expand the bracket"?
...
According to the question T becomes:

$T(D)=(16-\frac13 D) \cdot D^3$

Multiply completely that you get:

$T(D)=16 D^3 - \frac13 D^4$

which is much easier to differentiate then the term you tried to use

7. Originally Posted by earboth
According to the question T becomes:

$T(D)=(16-\frac13 D) \cdot D^3$

Multiply completely that you get:

$T(D)=16 D^3 - \frac13 D^4$

which is much easier to differentiate then the term you tried to use
Oh, I see it now.

So the derivative would be:
$48D^2-\frac43 D^3$?

8. Originally Posted by bpdude
...

So the derivative would be:
$48D^2-\frac43 D^3$?
Correct!

9. Originally Posted by earboth
Correct!
I see! Thank you, earboth!

Any clues about either 3 or 4?

I really appreciate the help guys! I'm glad I joined this forum, because it's certainly full of knowledgeable people!

10. Originally Posted by bpdude
3rd:
(describes a museum deciding to sell a painting and invest the proceeds)
basically:
"B(t)= P(t)(1.04)^(10-t)"

{where P(t)=sale price, B(t)=balance in yr. 2010, t=year when painting is sold, measured from the year 2000 so 0<t<10}

"Find B'(7), given that P(7)=120,000 and P'(7)=5,000.

B'(7)= _____

I have tried finding the derivative of the equation {B'(t)=P(t)ln(1.04)(1.04)^(10-t)}
and evaluating for t=7, but that didn't work. Any other ideas?
$
B(t) = P(t)1.04^{10-t}$

$B(7) = P(7)1.04^{10-7} = 120000 \times 1.04^3$

Use the product rule to differentiate:

$u = P(t) \: \rightarrow \: u' = P'(t)$

$v = 1.04^{10-t} \: \rightarrow \: v' = -1.04^{10-t} \cdot ln(1.04)$

Spoiler:
this is because v is of the form $a^x$ and $\frac{d}{dx}(a^x) = a^x\,ln(a)$. The minus comes from the chain rule

$\frac{d}{dt}[B(t)] = P'(t)\cdot 1.04^{10-t} - P(t)\cdot 1.04^{10-t}ln(1.04)$

Sub in t=7

Spoiler:
$B'(7) = 5000\cdot 1.04^{10-7} - 120000\cdot 1.04^{10-7}ln(1.04) = \ 330.16$

11. Originally Posted by e^(i*pi)
$
B(t) = P(t)1.04^{10-t}$

$B(7) = P(7)1.04^{10-7} = 120000 \times 1.04^3$

Use the product rule to differentiate:

$u = P(t) \: \rightarrow \: u' = P'(t)$

$v = 1.04^{10-t} \: \rightarrow \: v' = -1.04^{10-t} \cdot ln(1.04)$

Spoiler:
this is because v is of the form $a^x$ and $\frac{d}{dx}(a^x) = a^x\,ln(a)$. The minus comes from the chain rule

$\frac{d}{dt}[B(t)] = P'(t)\cdot 1.04^{10-t} - P(t)\cdot 1.04^{10-t}ln(1.04)$

Sub in t=7

Spoiler:
$B'(7) = 5000\cdot 1.04^{10-7} - 120000\cdot 1.04^{10-7}ln(1.04) = \ 330.16$
Nicely done, e^(i*pi)!

Just one more to go! You guys are awesome!

Those last 2 equations on the 4th question must be missing something, but I'm not sure what. Any ideas?

12. Any Ideas?

Anyone have anything yet? I'm stumped as to what it means by "...or state what additional information you would need to be able to compute the derivative". All the information looks like it's there to me, but apparently it isn't (since I've tried all the possible answers I could think of, and they didn't work).

13. Got it!

Never mind, I figured it out. My professor explained it to me during class this morning. I apparently had to use the product and power rules on (c) and (d) respectively. I guess that "..., or state what additional information..." part was throwing me off track.

I appreciate all of the help I received very much!
Thank you!