Question

\[f(x) = \dfrac{ax + b}{cx + d}\]Find the general form of\[f^{(n)}(x)\]

Answer 1

n'th Successive differentiation.

Tried Lebnitz rule??

ax+b and 1/(cx+d)

Answer 2

Please you have to use the induction principle

Answer 3

@ParthKohli

Answer 4

starting point:
\[\frac{d}{{dx}}\frac{{ax + b}}{{cx + d}} = \frac{{a\left( {cx + d} \right) - \left( {ax + b} \right)c}}{{{{\left( {cx + d} \right)}^2}}} = \frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\]

Answer 5

My book says that you can transform two functions into one and THEN use the induction principle, which makes things easier for us.

Answer 6

Here is what the transformation is:\[\dfrac{ax + b}{cx + d} = \dfrac{a}{c} + \dfrac{bc - ad}{c(cx + d)}\]

Answer 7

sincerely I have thought to the what you say, nevertheless if I apply repetedly the derivation operator I got your result anyway

Answer 8

oops... repeatedly...

Answer 9

for example I got these results:
\[\begin{gathered}
f''\left( x \right) = - 2c\frac{{ad - bc}}{{{{\left( {cx + d} \right)}^3}}} \hfill \\
f'''\left( x \right) = 6{c^2}\frac{{ad - bc}}{{{{\left( {cx + d} \right)}^4}}} \hfill \\
\end{gathered} \]

Answer 10

Yup, that's what I did to solve this question too. The reason I asked this is that transformation. Is that transformation useful in other areas of calculus? And does it have a special name?

Answer 11

yes! that transformation is useful in the area of the complex analysis, more precisely:
the transformation below:
\[w = f\left( z \right) = \frac{{az + b}}{{cz + d}}\]

Answer 12

where:
\[ad - bc \ne 0\]
establish a bijective correspondence between the z-plane and the w-plane

Answer 13

they are called "mappings"

Answer 14

or Moebius transformations

Answer 15

That's exactly the answer I was looking for! Thanks!

Answer 16

thanks!

Answer 17

you just did long division and calling it with crazy names haha

Answer 18

lol

Answer 19

I'm so stupid, haha...

Answer 20

Still laughing at my stupidity...

Answer 21

\[\begin{align}\dfrac{d^n}{dx^n}\left(\dfrac{ax+b}{cx+d}\right) &= \dfrac{d^n}{dx^n}\left(\frac{a}{c}+\frac{bc-ad}{c}\dfrac{1}{cx+d}\right)\\~\\
&= \dfrac{d^n}{dx^n}\left(\frac{a}{c}\right) + \frac{bc-ad}{c}\dfrac{d^n}{dx^n}\left(cx+d\right)^{-1}\\~\\
&= \dfrac{d^n}{dx^n}\left(\frac{a}{c}\right) + \frac{bc-ad}{c}\dfrac{(-1)^n(n!)c^n}{(cx+d)^{n+1}}\\~\\
\end{align}\]

Answer 22

Yeah, done...

Answer 23

In moving from second line to third line below formula is used
\[\dfrac{d^n}{dx^n}(ax+b)^m = a^nm(m-1)(m-2)\cdots(m-n+1)(ax+b)^{m-n}\]

plugin \(m=-1\)

Answer 24

Are you ganeshie?

Answer 25

idk who he is but this is not the first time somebody asking me this :/

Answer 26

WHAT THE HELL?!

I'VE ALWAYS BELIEVED THAT YOU'RE GANESHIE.

OK, so you're rsadhvika then?

Answer 27

thats another user that ppl confuse me and ganeshie wid hmm

Answer 28

No, I'm pretty sure you said somewhere that you're rsadhvika.

Answer 29

And I'm also pretty sure that you said somewhere that you're ganeshie.

Answer 30

By transitive property, ganeshie = rsadhvika! :O

Answer 31

ganeshie = rsadhvika
rational is different.