Question

Please explain the underlying logic -

Answer 1

fast cause that aint the way if you know what i mean

Answer 2

Consider an implicit function f(x,y) = c
Then,
\[\frac{ dy }{ dx }=-\frac{ f_x }{ f_y }\]
where f_x is differential coefficient of f(x,y) w.r.t. x ; treating y as constant.
Similarly, f_y is differentiation of f(x,y) w.r.t. y ; treating x as constant

For e.g.
\[f(x,y) = x^3 + y^3 - 3axy = 0\]
Then
\[f_x = 3x^2 - 3ay\]
\[f_y = 3y^2 - 3ax\]
\[\frac{ dy }{ dx }=-\frac{ x^2-ay }{ y^2-ax }\]

Answer 3

Can someone prove how
\[\frac{ dy }{ dx }=-\frac{ f_x }{ f_y }\]
??

Answer 4

@artofspeed
I am not good with partial differentiation. I can't see how
\[\frac{ \delta f }{ \delta x } \div \frac{ \delta f }{ \delta y } = -\frac{ dy }{ dx }\]

Answer 5

Remember that \[
\frac{df}{dx} = f_x\frac{dx}{dx}+f_y\frac{dy}{dx} = 0\implies f_y\frac{dy}{dx} =-f_x\implies
\frac{dy}{dx}=-\frac{f_x}{f_y}\]

Answer 6

When \(df/dx=0\)

Answer 7

@wio
\[\frac{ df }{ dx }=f_x \frac{ dx }{ dx } + f_y \frac{ dy }{ dx }\]

Let's say
\[h(t) = f(t) ^ {g(t)}\]

Can we use the following solution (in general) ??
Let f(t) = x ; g(t) = y
\[h'(t) = h_x \frac{ dx }{ dt } + h_y \frac{ dy }{ dt }\]

Answer 8

\[\frac{ dy }{ dx }=-\frac{ (\frac{ df }{ dx }) }{ (\frac{ df }{ dy}) }\]
\[\frac{ df }{ dy}\frac{ dy }{ dx }=-\frac{ df }{ dx }\]
f'=-f'
f'=0
f(x,y)=C since that's the only way you're going to get something to equal the negative of itself.

Answer 9

It's the chain rule.

Answer 10

@Kainui
I don't quite get your post.
f(x,y) = c does not imply
\[\frac{ df }{ dx }=0\]

or am I wrong ??

Answer 11

Okay, I got it.
I was wrong..

Answer 12

Thanxx for your time every1 :)

Answer 13

@LastDayWork \[
h(t) = f(t)^{g(t)} \implies \ln(h(t))=g(t)\ln(f(t))\implies \frac{h'(t)}{h(t)} = g'(t)\ln(f(t))+\frac{g(t)f'(t)}{f(t)}
\]This simplifies to\[
h'(t) = f(t)^{g(t)}g'(t)\ln(f(t))+g(t)f'(t)f(t)^{g(t)-1}
\]

Answer 14

One looks like the derivative of an exponential and the other looks like the derivative of a power function.

Answer 15

I know it works in this case. I was wondering if it works in general..
I can't see how is it chain rule..

Answer 16

It works if the functions are differentiable and defined.

Answer 17

Okay..thanxx for clarifying..

Answer 18

What do you mean by 'can't see how'?

Answer 19

The chain rule I have studied says
\[\frac{ dy }{ dx }=\frac{ dy }{ dt }\frac{ dt }{ dx }\]
I can't see terms denoting partial differentiation in this ^^

Answer 20

Consider \[
\frac{df(x,g(x))}{dx} = f_x(x,g(x))+f_g(x,g(x))g'(x)
\]

Answer 21

For \(f_x\) we treat \(g(x)\) as constant with respect to \(x\) and for \(f_g\) we treat \(x\) as constant with respect to \(x\).

Answer 22

Why are we adding the two terms (in RHS) ??
I mean, how do we justify it ?

Answer 23

The partial derivatives are orthogonal. Each one computes the difference in their dimension, so there is no overlap. You combine the differences through addition.

Answer 24

That's exactly what I was looking for..I wish I could give you more medals..XD