# Help with to derive this

• Aug 13th 2010, 03:59 PM
DoesNotCompute
Help with to derive this
Hi Guys,

this may seem like a very simple question to some of you.. but im struggling.. could someone please help me derive this:

f(x,y) = ln(x-3y)

Chain rule?.. if so ive confused myself with the x's and the y's

basicly ive been asked to find the linear approximation of that function at (7,2) and then use it to approximate f(6.9, 2.06).. could someone please confim the below process?

Find f(7,2)

Find f ′(x,y)

Find f ′(7,2)

Use formula for linear approximation near the point a .... f(x) ≈ f(a) + f ′(a) (x - a) not sure about this part.. what happened to the y's?

Substitute a = 7

then im unsure...

any help would be much appricated :-)
• Aug 13th 2010, 07:57 PM
mr fantastic
Quote:

Originally Posted by DoesNotCompute
Hi Guys,

this may seem like a very simple question to some of you.. but im struggling.. could someone please help me derive this:

f(x,y) = ln(x-3y)

Chain rule?.. if so ive confused myself with the x's and the y's

basicly ive been asked to find the linear approximation of that function at (7,2) and then use it to approximate f(6.9, 2.06).. could someone please confim the below process?

Find f(7,2)

Find f ′(x,y)

Find f ′(7,2)

Use formula for linear approximation near the point a .... f(x) ≈ f(a) + f ′(a) (x - a) not sure about this part.. what happened to the y's?

Substitute a = 7

then im unsure...

any help would be much appricated :-)

Do you know how to differentiate ln(x-a) wrt x? Do you know how to differentiate ln(b-3y) wrt y?

(You ought to if you're studying multivariable calculus. If you don't, then the question you have posted is not your immediate problem. Your immediate problem would be to urgently review all of the pre-requisite single variable calculus that your instructor will assume you already know).

Then you know how to calculate the partial derivatives $f_x$ and $f_y$.

Your linear approximation formula is the wrong one to use for this question. You have quoted the single variable formula. The formula required is the two variable formula. It will be in your classnotes or textbook. I suggest you find it (and then review all the examples that will follow it).
• Aug 13th 2010, 09:48 PM
DoesNotCompute
Quote:

Originally Posted by mr fantastic
Do you know how to differentiate ln(x-a) wrt x? Do you know how to differentiate ln(b-3y) wrt y?

(You ought to if you're studying multivariable calculus. If you don't, then the question you have posted is not your immediate problem. Your immediate problem would be to urgently review all of the pre-requisite single variable calculus that your instructor will assume you already know).

Then you know how to calculate the partial derivatives $f_x$ and $f_y$.

Your linear approximation formula is the wrong one to use for this question. You have quoted the single variable formula. The formula required is the two variable formula. It will be in your classnotes or textbook. I suggest you find it (and then review all the examples that will follow it).

Hi There,

you are correct in saying i have major gaps in my understanding of single variable and well as mutli. taking your advice i have looked through the book and its starting to make abit more sence now.. but im still confused as to how i differentiate the function... this is my delima =>

do i use:

dy/dx = dy/du . du/dx

so by chain rule

dy/du = -3/(x-3y) and du/dx = 1/(x-3y) so dy/dx = -3/(x-3y)

or

dx/dt = ∂f/∂x . dx/dt + ∂f/∂y . dy/dt ? ... what is "t"?
• Aug 13th 2010, 09:52 PM
mr fantastic
Quote:

Originally Posted by DoesNotCompute
Hi There,

you are correct in saying i have major gaps in my understanding of single variable and well as mutli. taking your advice i have looked through the book and its starting to make abit more sence now.. but im still confused as to how i differentiate the function... this is my delima =>

do i use:

dy/dx = dy/du . du/dx

so by chain rule

dy/du = -3/(x-3y) and du/dx = 1/(x-3y) so dy/dx = -3/(x-3y)

or

dx/dt = ∂f/∂x . dx/dt + ∂f/∂y . dy/dt ? ... what is "t"?

By the chain rule, $\frac{d}{dx} \ln (ax + b) = \frac{a}{ax + b}$. It can equally be said that $\frac{d}{dy} \ln (ay + b) = \frac{a}{ay + b}$.
• Aug 13th 2010, 10:25 PM
DoesNotCompute
thanks for your help again.. im sorry.. but im still confused:

chain rule: f'(x) = f '(g(x)) g '(x) so

f '(x) = 1/(x-3y) . 1
=> 1/(x-3y)

and then

f '(y) = 1/(x-3y) . -3
=> -3/(x-3y)

is this correct for the partial derivatives?
• Aug 14th 2010, 12:23 AM
mr fantastic
Quote:

Originally Posted by DoesNotCompute
thanks for your help again.. im sorry.. but im still confused:

chain rule: f'(x) = f '(g(x)) g '(x) so

f '(x) = 1/(x-3y) . 1
=> 1/(x-3y)

and then

f '(y) = 1/(x-3y) . -3
=> -3/(x-3y)

is this correct for the partial derivatives?

Overlooking the poor notation, your answers for the partial derivatives are correct.