what's f'(x,y) when f(x,y) = g(x-2y), g'(1)=3

[helenli89]

I have come cross this question:

A function f: R2 -> R is defined by f(x,y) = g(x-2y), where g: R -> R. If g'(1) = 3, calculate fx(3,1) and fy(3,1). Show that fy(x,y) = -2 * fx(x,y) in general.

Let q(x) = x-2y. So, I think that f(x,y) = g(x-2y) as f(x,y) = g(q(x)), so f'(x,y) = g'(q(x)) * q'(x).
so, fx(x,y) = g'(q(x)) * (1-2y) and fy(x,y) = g'(q(x)) * (x-2) then fx(3,1) = -3 and fy(x,y) = 3.
But looking at the prove of fy(x,y) = -2 * fx(x,y) I think my answer for the fy(3,1) is wrong. Could someone please help me?

ps. I was just wondering is there any function that is biger then ln?
Thanks.

Helen

 

[DrMike]

I think that f(x,y) = g(x-2y) as f(x,y) = g(q(x))

In fact, q is not a function of just x, but of x and y. It would be better to write f(x,y) = g(q(x,y))

so f'(x,y) = g'(q(x)) * q'(x)

You're on the right track, but more accurately, f[sub:1bqisc1a]w[/sub:1bqisc1a](x,y)=g'(q(x,y))*q[sub:1bqisc1a]w[/sub:1bqisc1a](x,y), where w is either x or y.

f[sub:1bqisc1a]x[/sub:1bqisc1a](x,y) = g'(q(x)) * (1-2y) and f[sub:1bqisc1a]y[/sub:1bqisc1a](x,y) = g'(q(x)) * (x-2)

You've got your derivatives of q wrong here, that's messing things up later. I'll go through an example step by step :

Let h(x,y) = 1 + x + 3y + x[sup:1bqisc1a]2[/sup:1bqisc1a]y. First, I'll differentiate it with respect to x, treating y as a constant.

* The derivative of a sum is the sum of teh derivatives, so I can differentiate each term separately.
* 1 is a constant, so the derivative is 0.
* the derivative of x, with respect to x, is 1.
* 3y is.. well, I'm treating y as a constant - it doesn't change when x changes - so 3y also doesn't change with x. The derivative of 3y with respect to x is 0.
* Again, since y is treated as a constant here, the derivative of yx[sup:1bqisc1a]2[/sup:1bqisc1a] is no harder than, say, Ax[sup:1bqisc1a]2[/sup:1bqisc1a]. I get 2yx.

Therefore, h[sub:1bqisc1a]x[/sub:1bqisc1a](x,y)=0+1+0+2yx = 1+2xy.

Now, I'll differentiate with respect to y, treating x as a constant (ie, not changing with y).

* 1 is a constant, so the derivative is 0.
* x is also constant, at least, with respect to y. It doesn't change as y changes. The derivative is 0.
* The derivative of 3y is 3.
* Again, since x is treated as a constant here, the derivative of x[sup:1bqisc1a]2[/sup:1bqisc1a]y with respect to y is just x[sup:1bqisc1a]2[/sup:1bqisc1a].

Therefore, h[sub:1bqisc1a]x[/sub:1bqisc1a](x,y)=0+0+3+x[sup:1bqisc1a]2[/sup:1bqisc1a] = 3+x[sup:1bqisc1a]2[/sup:1bqisc1a].

Now try this with your q(x,y)=x-2y.

ps. I was just wondering is there any function that is biger then ln?

You can always make a function bigger than any given function, eg, ln(x)+1 will be bigger than ln(x)...

But what do you mean, exactly?

 

[helenli89]

ps. I was just wondering is there any function that is biger then ln?

You can always make a function bigger than any given function, eg, ln(x)+1 will be bigger than ln(x)...

But what do you mean, exactly?[/quote]

First of all thank you very much for reply my posting.
Secondly, what I mean here is that is there any x^n functions that is bigger than ln(x), ie doesn x^8, for example, bigger than ln(x)?
Thanks.

 

[helenli89]

I also checked, in maple software that d(x^2y + xy)/dx it equal to 2xy which is tha same as d(x^2y+y)/dx. can someone tell why?

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ f(x,y) \ = \ g(x-2y), \ g \ ' \ (1) \ = \ 3\)

\(\displaystyle Find \ f \ ' \ (x,y), \ f_x(3,1), \ f_y(3,1), \ and \ show \ f_y(x,y) \ = \ -2[f_x(x,y)]\)

\(\displaystyle If \ anyone \ can \ solve \ this, \ I \ would \ be \ interested \ in \ seeing \ your \ solution, \ not \ your \ happy \ horseshit.\)