# An Example on Differentials ... Browder Proposition 8.12 ...

#### Peter

##### Well-known member
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I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some help with an example based only on Proposition 8.12 ... and the material in Section 8.2 preliminary to Proposition 8.12 (see scanned text at end of post) ... namely Definitions 8.9 and 8.10 and Proposition 8.11 ...

For Definitions 8.9 and 8.10 and Proposition 8.11 ... see scanned text below ...

Now consider the following example ...

$$\displaystyle f(x,y) = (xy, x/y)$$ where we have $$\displaystyle f_1(x, y) = xy$$ and $$\displaystyle f_2(x, y)= x/y$$

I wish to determine $$\displaystyle L = \text{df}_p$$ where $$\displaystyle p = (a, b)$$ using only Proposition 8.12 and the material in Section 8.2 preliminary to Proposition 8.12 (see scanned text at end of post) ... namely Definitions 8.9 and 8.10 and Proposition 8.11 ...

I also wish to determine the Jacobian matrix of $$\displaystyle f$$ at $$\displaystyle p$$, namely, $$\displaystyle f'(p)$$ again using only Proposition 8.12 and the material in Section 8.2 preliminary to Proposition 8.12 (see scanned text at end of post) ... namely Definitions 8.9 and 8.10 and Proposition 8.11 ...

I am unable to carry out the above without using partial derivatives ... which Browder has not introduced yet... indeed, it may not be possible ...

Hope someone can help ...

Peter

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Browder Section 8.2 (preliminary to Proposition 8.12) reads as follows:

Hope that helps ...

Peter

#### HallsofIvy

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That looks straight forward. You are given f(x,y)= (xy, x/y). The partial derivative of xy with respect to x is y, the partial derivative of xy with respect to x is y, the partial derivative of x/y with respect to x is 1/y, and the partial derivative of x/y with respect to y is $$-x/y^2$$. So $$df= (ydx+ xdy, dx/y- xdy/y^2)$$.
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The Jacobian is $$\begin{bmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{bmatrix}= \begin{bmatrix} y & x \\ \frac{1}{y} & -\frac{x}{y^2}\end{bmatrix}$$.

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#### Peter

##### Well-known member
MHB Site Helper
That looks straight forward. You are given f(x,y)= (xy, x/y). The partial derivative of xy with respect to x is y, the partial derivative of xy with respect to x is y, the partial derivative of x/y with respect to x is 1/y, and the partial derivative of x/y with respect to y is $$-x/y^2$$. So $$df= (ydx+ xdy, dx/y- xdy/y^2)$$.
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The Jacobian is $$\begin{bmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{bmatrix}= \begin{bmatrix} y & x \\ \frac{1}{y} & -\frac{x}{y^2}\end{bmatrix}$$.

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Thanks HallsofIvy ... but ...

My questions were regarding a process or method to determine $$\displaystyle L = \text{df}_p$$ where $$\displaystyle p = (a, b)$$ using only Proposition 8.12 and the material in Section 8.2 preliminary to Proposition 8.12 (see scanned text at end of post) ... namely Definitions 8.9 and 8.10 and Proposition 8.11 ... this does not include using partial differentiation as a process

... ... and similarly I wished to know if there is a method or process to determine the Jacobian matrix of $$\displaystyle f$$ at $$\displaystyle p$$, namely, $$\displaystyle f'(p)$$ again using only Proposition 8.12 and the material in Section 8.2 preliminary to Proposition 8.12 (see scanned text at end of post) ... namely Definitions 8.9 and 8.10 and Proposition 8.11 ...

Maybe Proposition 8.12 only states some theoretical relationships concerning $$\displaystyle L = \text{df}_p$$ and $$\displaystyle f'(p)$$ and does not concern itself with how to determine or calculate these entities ...

Nonetheless ... thank you again for your post ... it is helpful to me as although i have no trouble determining the Jacobian matrix of $$\displaystyle f$$ at $$\displaystyle p$$, namely, $$\displaystyle f'(p)$$ ... I do need help in determining the associated linear transformation $$\displaystyle L = \text{df}_p$$ where $$\displaystyle p = (a, b)$$ ...

I'll use a mix of your notation and Browder's to attempt to determine and illustrate the linear transformation $$\displaystyle L = \text{df}_p$$ where $$\displaystyle p = (a, b)$$ ...

We have

$$\displaystyle f(x,y) = (xy, x/y)$$ where we have $$\displaystyle f_1(x, y) = xy$$ and $$\displaystyle f_2(x, y)= x/y$$

and we have

$$\displaystyle \text{df}_p (h) = ( \text{df}_{ 1 (p) } (h) , \text{df}_{ 2 (p) } (h) )$$

where $$\displaystyle h = (h_1, h_2) = ( \text{dx}, \text{dy} )$$

Now we also have

$$\displaystyle \text{df}_{ 1 (p) } (h) = ( \frac{\partial f_1}{\partial x} , \frac{\partial f_1}{\partial y} ) \begin{pmatrix} \text{dx} \\ \text{dy} \end{pmatrix} \ |_{ p = (a, b) }$$

$$\displaystyle \frac{\partial f_1}{\partial x} \text{dx} + \frac{\partial f_1}{\partial y} \text{dy} \ |_{ p = (a, b) }$$

$$\displaystyle = y \text{dx} + x \text{dy} \ |_{ p = (a, b) }$$

$$\displaystyle = y h_1 + x h_2 \ |_{ p = (a, b) }$$

$$\displaystyle = b h_1 + a h_2$$

... and also

$$\displaystyle \text{df}_{ 2 (p) } (h) = ( \frac{\partial f_2}{\partial x} , \frac{\partial f_2}{\partial y} ) \begin{pmatrix} \text{dx} \\ \text{dy} \end{pmatrix} \ |_{ p = (a, b) }$$

$$\displaystyle = \frac{\partial f_2}{\partial x} \text{dx} + \frac{\partial f_2}{\partial y} \text{dy} \ |_{ p = (a, b) }$$

$$\displaystyle = \frac{ \text{dx} }{ y } - \frac{ x \text{dy} }{ y^2 } \ |_{ p = (a, b) }$$

$$\displaystyle = \frac{ h_1 }{ y } - \frac{ x h_2 }{ y^2 } \ |_{ p = (a, b) }$$

$$\displaystyle = \frac{ h_1 }{ b } - \frac{ a h_2 }{ b^2 }$$

Is that correct?

Now ... and I am even more unsure here ...

... if we want to determine $$\displaystyle L = \text{df}_p = ( \text{df}_{ 1 (p) } , \text{df}_{ 2 (p) }$$ ...

then we have ...

$$\displaystyle L = \text{df}_p = ( (y, x) , ( \frac{1}{y} , \frac{ -x}{ y^2} ) |_{ p = (a, b) }$$

$$\displaystyle = ( (b, a), ( \frac{1}{b} , \frac{ -a}{ b^2} )$$

Is that correct?

Hope someone can comment on and critique the above interpretation of Browder's terminology ...

Peter

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