How to find the partial derivative of this using the definition?

**AlphaRock** · April 6th 2010, 08:27 PM

f(x,y) = 2(x^2)y - 4x/y

Prove that f'y (with respect to y) = 2x^2 + 4x/(y^2) using the definition of a partial derivative (limit)

**Prove It** · April 6th 2010, 08:34 PM

Originally Posted by AlphaRock

f(x,y) = 2(x^2)y - 4x/y

Prove that f'y (with respect to y) = 2x^2 + 4x/(y^2) using the definition of a partial derivative (limit)

To take the derivative with respect to $y$ , you treat $x$ as a constant and use the standard derivative rules.

$f(x, y) = 2x^2y - \frac{4x}{y}$

$= 2x^2y^1 - 4x\,y^{-1}$ .

$\frac{\partial f}{\partial y} = 1\cdot 2x^2 y^{1-1} - 1(-4x)y^{-1-1}$

$= 2x^2y^0 + 4x\,y^{-2}$

$= 2x + \frac{4x}{y^2}$ .

**AlphaRock** · April 6th 2010, 08:36 PM

Originally Posted by Prove It

To take the derivative with respect to $y$ , you treat $x$ as a constant and use the standard derivative rules.

$f(x, y) = 2x^2y - \frac{4x}{y}$

$= 2x^2y^1 - 4x\,y^{-1}$ .

$\frac{\partial f}{\partial y} = 1\cdot 2x^2 y^{1-1} - 1(-4x)y^{-1-1}$

$= 2x^2y^0 + 4x\,y^{-2}$

$= 2x + \frac{4x}{y^2}$ .

Thanks for your help, Prove it.

How would we find f'y using limits?
The definition of this partial derivative would be:

lim (f(x,y+k) - f(x,y))/k
k->0

**undefined** · April 6th 2010, 09:20 PM

Originally Posted by AlphaRock

Thanks for your help, Prove it.

How would we find f'y using limits?
The definition of this partial derivative would be:

lim (f(x,y+k) - f(x,y))/k
k->0

The calculation is tedious but straightforward.

$\frac{\partial f}{\partial y} = \lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}$

$=\lim_{h\to 0}\frac{2x^2(y+h)-\frac{4x}{y+h}-(2x^2y-\frac{4x}{y})}{h}$

$=\lim_{h\to 0}\frac{2hx^2-\frac{4x}{y+h}+\frac{4x}{y}}{h}$

$=\lim_{h\to 0}\bigg(2x^2+\frac{1}{h}\bigg(\frac{-4x(y)}{y(y+h)}+\frac{4x(y+h)}{y(y+h)}\bigg)\bigg)$

$=\lim_{h\to 0}\bigg(2x^2+\frac{-4xy+4xy+4xh}{h(y)(y+h)}\bigg)$

$=\lim_{h\to 0}\bigg(2x^2+\frac{4x}{(y)(y+h)}\bigg)$

$=2x^2+\frac{4x}{(y)(y+0)}$

$=2x^2+\frac{4x}{y^2}$

**Prove It** · April 6th 2010, 09:23 PM

$\frac{f(x, y + h) - f(x, y)}{h}$

$= \frac{2x^2(y + h) - \frac{4x}{y + h} - \left(2x^2y - \frac{4x}{y}\right)}{h}$

$= \frac{2x^2h + \frac{4xh}{y(y + h)}}{h}$

$= \frac{\frac{2x^2h\,y(y + h) + 4h}{y(y + h)}}{h}$

$= \frac{2x^2h\,y(y + h) + 4xh}{h\,y(y + h)}$

$= \frac{h[2x^2y(y + h) + 4x]}{h\,y(y + h)}$

$= \frac{2x^2y(y + h) + 4x}{y(y + h)}$ .

Now making $h \to 0$ we get

$\frac{2x^2y^2 + 4x}{y^2}$

$= 2x^2 + \frac{4x}{y^2}$ .

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