1. ## Partial Differentiation

$f(x,y)=2x^3y^2+2y+4x$. Find $f_x(1,3)$ (that is, use the definition to find the partial derivative $f_x(x,y)$ at (1,3)).

The answer has to be $f_x(1,3) = 58$.

My Attempt:

I want to use the following definition to find the partial derivative

$f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0}$ $= \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}$

So,

$\lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}$

$\lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}$

If I cancel out $\Delta x$ in the denominator with the the term on top I get:

$\lim_{\Delta x \to 0} 10x-52$

which is wrong... can anoyne help?

2. Originally Posted by demode
$f(x,y)=2x^3y^2+2y+4x$. Find $f_x(1,3)$ (that is, use the definition to find the partial derivative $f_x(x,y)$ at (1,3)).

The answer has to be $f_x(1,3) = 58$.

My Attempt:

I want to use the following definition to find the partial derivative

$f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0}$ $= \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}$

So,

$\lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}$

$\lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}$

If I cancel out $\Delta x$ in the denominator with the the term on top I get:

$\lim_{\Delta x \to 0} 10x-52$

which is wrong... can anoyne help?
You just differentiate each term with respect to $x$, keeping $y$ constant.

So $\frac{\partial f}{\partial x} = 6x^2y^2 + 4$.

Now evaluate this at $(x, y) = (1, 3)$.

3. Originally Posted by demode
$f(x,y)=2x^3y^2+2y+4x$. Find $f_x(1,3)$ (that is, use the definition to find the partial derivative $f_x(x,y)$ at (1,3)).

The answer has to be $f_x(1,3) = 58$.

My Attempt:

I want to use the following definition to find the partial derivative

$f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0}$ $= \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}$
Where did the '64' come from? f(1, 3)= 2(1)(9)+ 2(3)+ 4(1)= 18+ 6+ 4= 28.

And does "from the definition" really mean using the difference quotient? By the time you get to partial derivatives, you usually use the definition that Prove It refers to- that the partial derivative of f with respect to x is just the ordinary derivative holding other variables constant.

So,

$\lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}$
No, this should be
$\frac{2(1+ \Delta x)^3(3^2)+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}$

[/quote] $\lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}$

If I cancel out $\Delta x$ in the denominator with the the term on top I get:

$\lim_{\Delta x \to 0} 10x-52$

which is wrong... can anoyne help?[/QUOTE]

4. $\lim_{\Delta x \to 0} \frac{2(1+ \Delta x)^3(3^2)+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}$

$\lim_{\Delta x \to 0} \frac{2(1+3(\Delta x)+3(\Delta x)^2+(\Delta x)^3)9+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}$

$\lim_{\Delta x \to 0} \frac{18+54(\Delta x)+54(\Delta x)^2 + 18 (\Delta x)^3 +10+ 4 (\Delta x)- 28}{\Delta x}$

$
\lim_{\Delta x \to 0} \frac{18(\Delta x)^3+54(\Delta x)^2+58(\Delta x)}{\Delta x}
$

So, how exactly do I need to evaluate this limit to get $58$?

Can I cancel out the $\Delta x$ on top with the $\Delta x$ on the denominator to get

$\lim_{\Delta x \to 0} 18 (\Delta x)^2 +54(\Delta x) +58$

And then substitute 0's

$18 (0)^2 +54(0) +58 = 58$

Is this correct?