[SOLVED] Chain rule with functions depending on other functions

**garymarkhov** · August 12th 2009, 03:43 AM

Hello all! I'm stuck in a big way on this one.

Here are the givens:

q1(P,S,y) = 6*P^(-2)*S^(3/2)*y

q2(P,S,y) = 4*P*S^(-1)*y^2

P(t) = (12*t)^(1/2)
S(r,t) = 10*r*t^2
y(r) = 20r.

Find the derivative of q1 and q2 with respect to both time t and interest rate r when t=3 and r=0.1.

------------------------------

Here's my methodology:

partial q1/partial t = partial q1/partial P*dP/dt + partial q1/partial S*partial S/partial t + partial q1/partial y*dy/dt

partial q1/partial r = partial q1/partial P*dP/dr + partial q1/partial S*partial S/partial r + partial q1/partial y*dy/dr

partial q2/partial t = partial q2/partial P*dP/dt + partial q2/partial S*partial S/partial t + partial q2/partial y*dy/dt

partial q2/partial r = partial q2/partial P*dP/dr + partial q2/partial S* partial S/partial r + partial q2/partial y*dy/dr

------------------------------

Getting started on the calculations, I plugged in (P=6, S=9, y=2) to the partials of q1 and q2:

partial q1/partial P = -12*P^(-3)*S^(3/2)*y = -3

partial q1/partial S = 9*P^(-2)*S^(1/2)*y = 1.5

partial q1/partial y = 6*P^(-2)*S^(3/2) = 4.5

partial q2/partial P = 4*S^(-1)*y^2 = 16/9

partial q2/partial S = -4*P*S^(-2)*y^2 = -32/27

partial q2/partial y = 8*P*S^(-1)*y = 32/3

Putting those numbers into a matrix:

Row 1 (q1): -3 1.5 4.5
Row 2 (q2): 16/9 -32/27 32/3

Then putting the derivatives of the functions upon which the first functions depend, evaluated at t=3 and r=0.1 into a matrix (the derivative wrt t is in the first column, and the derivative wrt r is in the second column):

Row1: 1 0
Row 2: 6 90
Row 3: 0 20

Multiplying my two matrices together, I get:

Row 1: 6 225
Row 2: -5.33 106.66

But none of those numbers are correct (unless there is a mistake in the text I am using). Can someone please point out where I'm going wrong?

Thanks and all the best!

**alunw** · August 12th 2009, 06:27 AM

As nobody else seems to be helping you yet you might like to know that I've actually tried all the calculations myself now and I get what you get, and at least the dq2 by dr value matches what you get when you try by the method I suggested in your first set of posts.
I've also convinced myself your method is OK by trying it on various other functions.
There's still a misprint - you've got the sign wrong for dq2/dy in your initial list of partial derivatives though it is correct in your matrix.
So I think either the solution in your book is wrong, or possibly that you have misread the problem somehow or other - perhaps by getting one of the functions wrong somehow. Let's hope someone else can shed some light on the problem.

**Amer** · August 12th 2009, 06:54 AM

Originally Posted by garymarkhov

Hello all! I'm stuck in a big way on this one.

Here are the givens:

q1(P,S,y) = 6*P^(-2)*S^(3/2)*y

q2(P,S,y) = 4*P*S^(-1)*y^2

P(t) = (12*t)^(1/2)
S(r,t) = 10*r*t^2
y(r) = 20r.

Find the derivative of q1 and q2 with respect to both time t and interest rate r when t=3 and r=0.1.

------------------------------

Thanks and all the best!

$q_1(p,s,y)=6p^{-2}(t)[s^{\frac{3}{2}}(r,t)y(r)]$

$q_2(p,s,y)=4p(t)[s^{-1}(r,t)y^2(r)]$

$p(t)=(12t)^{\frac{1}{2}}$

$s(r,t)=10r(t)^2$

$y(r)=20r$

you can sub p and s and y values

$q_1=6((12t)^{\frac{1}{2}})^{-2})[10r(t)^2]^{\frac{3}{2}}(20r)$

$q_1=6(12t)^{-1}[(10r)^{\frac{3}{2}}t^3](20r)$

$q_1(t,r)=10^{\frac{5}{2}}r^{\frac{5}{2}}t^2$ it is now more easy

$q_1'(t,r)=10^{\frac{5}{2}}\left[\left(2t(r^{\frac{5}{2}})\right)+t^2\left(\frac{5} {2}(r)^{\frac{3}{2}}\right)\right]$

$q_1'(3,0.1)=10^{\frac{5}{2}}\left[\left(2(3)(0.1^{\frac{5}{2}})\right)+3^2\left(\fra c{5}{2}(0.1)^{\frac{3}{2}}\right)\right]$

**garymarkhov** · August 12th 2009, 06:55 AM

Originally Posted by alunw

As nobody else seems to be helping you yet you might like to know that I've actually tried all the calculations myself now and I get what you get, and at least the dq2 by dr value matches what you get when you try by the method I suggested in your first set of posts.
I've also convinced myself your method is OK by trying it on various other functions.
There's still a misprint - you've got the sign wrong for dq2/dy in your initial list of partial derivatives though it is correct in your matrix.
So I think either the solution in your book is wrong, or possibly that you have misread the problem somehow or other - perhaps by getting one of the functions wrong somehow. Let's hope someone else can shed some light on the problem.

As always, thanks for your eagle eyes. I've fixed the error. I can't guarantee that there aren't a few more errors (that's why I need help!), but I can say that I've double checked the initial problem setup several times to make sure that I'm not hamstringing you guys from the get-go.

By the way, if you know of similar problems I can practice on, I would love to see them. Let me know.

mr fantastic, care to weigh in? Anyone else?

**garymarkhov** · August 12th 2009, 07:22 AM

Originally Posted by Amer

$q_1(p,s,y)=6p^{-2}(t)[s^{\frac{3}{2}}(r,t)y(r)]$

$q_2(p,s,y)=4p(t)[s^{-1}(r,t)y^2(r)]$

$p(t)=(12t)^{\frac{1}{2}}$

$s(r,t)=10r(t)^2$

$y(r)=20r$

you can sub p and s and y values

$q_1=6((12t)^{\frac{1}{2}})^{-2})[10r(t)^2]^{\frac{3}{2}}(20r)$

$q_1=6(12t)^{-1}[(10r)^{\frac{3}{2}}t^3](20r)$

$q_1(t,r)=10^{\frac{5}{2}}r^{\frac{5}{2}}t^2$ it is now more easy

$q_1'(t,r)=10^{\frac{5}{2}}\left[\left(2t(r^{\frac{5}{2}})\right)+t^2\left(\frac{5} {2}(r)^{\frac{3}{2}}\right)\right]$

$q_1'(3,0.1)=10^{\frac{5}{2}}\left[\left(2(3)(0.1^{\frac{5}{2}})\right)+3^2\left(\fra c{5}{2}(0.1)^{\frac{3}{2}}\right)\right]$

Thanks Amer! I agree, that does make it a bit simpler (although I'm not bothered by the complexity of the method - I just want to know why I'm seemingly getting the wrong answer). I'm impressed by your Latex skills, btw.

Using your method, I get partial P/partial r = 225 and partial P/partial t = 6. Same as in my final matrix. This is starting to look bad for either me (I can't properly write down what's in my book) or my book's answer key.

Anyone else?

**Amer** · August 12th 2009, 07:29 AM

second one in other way

$q_1=4ps^{-1}y^2$

$\nabla q_1 = 4\left(\frac{\partial q}{\partial p} \frac{dp}{dt}\right)\left[s^{-1}y^2\right]+4 \frac{\partial q}{\partial s} \left(\frac{\partial s}{\partial t}+\frac{\partial s}{\partial r}\right)p(y^2)+$ $4\left(\frac{\partial q}{\partial y}\frac{dy}{dr}\right)\left[p(s)^{-1}\right]$

$\nabla q_1 = 4\left(12\frac{(12t)^{\frac{-1}{2}}}{2} \right)\left[s^{-1}y^2\right]+$ $4 (-s^{-2}) \left(20r(t)+10t^2\right)p(y^2)+$ $4\left(2y(20)\right)\left[p(s)^{-1}\right]$

$\nabla q_1 = 4\left(6(12t)^{\frac{-1}{2}} \right)\left[s^{-1}y^2\right]+4 (-s^{-2}) \left(20r(t)+10t^2\right)p(y^2)+$ $4\left(40y\right)\left[p(s)^{-1}\right]$

$p(3)=(12(3))^{\frac{1}{2}}=6$

$s(0.1,3)=10(0.1)(3)3^2=9$

$y(0.1)=20(0.1)=2$

$\nabla q_1 = 4\left(6(12(3))^{\frac{-1}{2}} \right)\left[9^{-1}2^2\right]+4 (-(9)^{-2}) \left(20(0.1)(3)+10(3)^2\right)6(2^2)+$ $4\left(40(2)\right)\left[6(9)^{-1}\right]$

$\nabla q_1 = 4\left(6(12(3))^{\frac{-1}{2}} \right)\left[9^{-1}2^2\right]+4 (-(9)^{-2}) \left(20(0.1)(3)+10(3)^2\right)6(2^2)+$ $4\left(40(2)\right)\left[6(9)^{-1}\right]$

$\nabla q_1 =4\left(6\frac{1}{6}\right)\left(\frac{4}{9}\right )+\frac{-4}{9}\left(6+10(9)\right)(24)+(160)\left(\frac{6}{ 9}\right)$

I wish I dose not make mistakes

**Amer** · August 12th 2009, 08:56 AM

what is the correct answer ??

I solve the first one and it is the same as your solution I get 231

**garymarkhov** · August 12th 2009, 04:41 PM

Originally Posted by Amer

what is the correct answer ??

I solve the first one and it is the same as your solution I get 231

My text, Mathematics for Economists (Simon), claims that the solution is

partial q1/partial r = 427.5

partial q1/partial t = 19.5

partial q2/partial r = -400/3

partial q2/partial t = -20/3

None of those answers are even close to what we're coming up with, are they?

Thread: [SOLVED] Chain rule with functions depending on other functions

LinkBack

Thread Tools

Display

[SOLVED] Chain rule with functions depending on other functions

Similar Math Help Forum Discussions

Using the Chain Rule for Complex Functions

Chain rule for exponential functions

[SOLVED] chain rule with functions related by equations?

the chain rule using trigonometric functions

functions using chain rule

Search Tags