# Math Help - [SOLVED] Chain rule with functions depending on other functions

1. ## [SOLVED] Chain rule with functions depending on other functions

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.

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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

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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!

2. 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.

3. 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

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

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

4. 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?

5. 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?

6. 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

7. what is the correct answer ??

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

8. 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?