Question

Help with calculus.

Find the n th derivative of f(x) = (ax+b)/(cx+d)

By the way, i can get a (in some sense) "recursive" formula involving the previous derivatives, but we're asked for an explicit formula. Thanks

Answer 1

http://en.wikipedia.org/wiki/General_Leibniz_rule

where g is g^(-1)

Answer 2

$$f(x)=\frac{ax+b}{cx+d}=\frac1c\frac{ax+b}{x+d}=\frac1c\frac{a(x+d)+(b-ad)}{x+d}=\frac1c\left(a+\frac{b-ad}{x+d}\right)$$

Answer 3

so we see that $$f'(x)=\frac{b-ad}c\cdot\frac{-1}{(x+d)^2}\\f''(x)=\frac{b-ad}c\cdot\frac{(-1)(-2)}{(x+d)^3}\\\dots\\f^{(n)}(x)=\frac{b-ad}c\cdot\frac{(-1)^nn!}{(x+d)^{n+1}}$$

Answer 4

oops that's actually slightly broken... just replace \(d\) in the stuff with \(d/c\)

Answer 5

I got it, the trick was to decompose the fraction.

Answer 6

$$f(x)=\frac{ax+b}{cx+d}=\frac1c\frac{ax+b}{x+d/c}=\frac1c\frac{a(x+d/c)+(b-ad/c)}{x+d/c}=\frac1c\left(a+\frac{b-ad/c}{x+d/c}\right)$$

Answer 7

$$f'(x)=\frac{b-ad/c}c\cdot\frac{-1}{(x+d/c)^2}\\f''(x)=\frac{b-ad/c}c\cdot\frac{(-1)(-2)}{(x+d/c)^3}\\\dots\\f^{(n)}(x)=\frac{b-ad/c}c\cdot\frac{(-1)^nn!}{(x+d/c)^{n+1}}$$

Answer 8

note:$$\frac{b-ad/c}c=\frac{bc-ad}{c^2}=(ad-bc)\cdot\frac{-1}{c^2}$$

Answer 9

so our answer is $$f^{(n)}(x)=(ad-bc)\frac{(-1)^{n+1}n!}{c^2(x+d/c)^{n+1}}$$

Answer 10

Ok, thanks

Answer 11

$$f^{(n)}(x)=(ad-bc)c^{n-1}n!\left(-\frac1{cx+d}\right)^{n+1}$$

Answer 12

just making it neater

Answer 13

I multiplied top and bottom by \(c^{n+1}\) and then made \(c^{n+1}(x+d/c)^{n+1}=(cx+d)^{n+1}\) and of course combined the \((-1)^{n+1}/(cx+d)^{n+1}=(-1/(cx+d))^{n+1}\)