Question

find the derivative of x^y=y^x with respect to x

Answer 1

@ranga

Answer 2

\[\Large x^y\quad=\quad y^x\]Taking natural log of each side:\[\Large \ln x^y\quad=\quad \ln y^x\]Applying rule of logs:\[\Large y \ln x\quad=\quad x \ln y\]Understand up to that point?
From here, we can differentiate implicitly.

Answer 3

yup i got it up to that far

Answer 4

Hmm so looks like we need product rule on each side:\[\Large \color{#F35633}{(y)'}\ln x+ y\color{#F35633}{(\ln x)'}\quad=\quad \color{#F35633}{(x)'}\ln y+ x\color{#F35633}{(\ln y)'}\]

Answer 5

ok. then what

Answer 6

So we need to take each of these derivatives.
Any of them look rather confusing?
Do you understand what the derivative of ln y will give us?

Answer 7

x'=1? and (lny)'=1/y'?

Answer 8

\[\Large (\ln x)'\quad=\quad \frac{1}{x}(x)'\quad=\quad \frac{1}{x}(1)\]

Answer 9

\[\Large (\ln y)'\quad=\quad \frac{1}{y}(y)'\quad=\quad \frac{1}{y}y'\]

Answer 10

oh ok then i should be fine. if i have any questions i will ask thanks

Answer 11

k cool :)

Answer 12

is it \[(\ln y-(y/x))/(\ln x-(x/y))\]

Answer 13

@zepdrix

Answer 14

Hmm ya that's fine.
It can certainly be simplified though :O
So we don't have those ugly fractions within a fraction.