Question

derivative of f(x) = tan^-1(kx) is
k/(1+k^2*x^2)
if theta is tan^-1(y/x) find dtheta/dy

Answer 1

\[f(x) = \tan^{-1} kx\]
derative is \[\frac{ k }{ 1+x^2k^2 }\]
if \[\phi = \tan^{-1} \frac{ y }{ x }\]
find \[\frac{ d \phi }{ dy}\]

Answer 2

tan^-1(y*x^-1) I have now moved up x
and it becomes negative. Because Iäm dervating for dy, I know x^-1 can be written as a constant.

Answer 3

Since its the same formula as the first one I write
k = x^-1 or k = 1/x
so \[\frac{ k }{ 1+x^2*k^2 }\]
can be written as \[\frac{ 1/x }{ 1+((1/x)^2)*ÿ^2}\]
Notice as k is replaced by 1/x, x is replaced by y
making x^2 = y^2.

Answer 4

what should be my next step?

Answer 5

@phi

Answer 6

@hartnn

Answer 7

\[\frac{ x }{ x^2+y^2 }\] is the correct answer, I'm at the final step but when I shorten it down it goes wrong..

Answer 8

nothing much to simplify
just multiply numerator and denominator by x^2
and you will get the required answer

Answer 9

(1/x)*x^2 = x = numerator

Answer 10

y^2/x^2 * x^2 = y^2 only

Answer 11

but the correct answer is x /(x^2 + y^2)

Answer 12

oh Im so stupid, I solved it, or we both did:)

Answer 13

yes