Question

Can someone explain this to me? Derivative of an inverse function:

Answer 1

Find \(f^{-1}'\) (d) for the given \(f\) and \(d\)

\(f(x)=\sqrt{3x+1}\), \(d=1\)

Answer 2

let y = f(x), then f(y)=x is its inverse

dy/dx = f' and dx/dy = (f^-1)' which is just 1/(dy/dx)

Answer 3

i dont know what (d) is in this context tho ... maybe the first derivative? or maybe x=d=1??

Answer 4

I'm given this too:

If \(c\) is a value of \(x\), in the interval \([a, b]\) and \(f(c)=d\), then \(f^{-1}'(d)\)=\(\frac{1}{f'(c)}\)

I'm assuming you solve for d.. our teacher threw this at us just last night.

Answer 5

then d is a value for x ... and yeah, that states the same thing

invertible functions have reciprocal rates of change.

Answer 6

of f and f^-1 are inverses of each other such that:

f(f^-1(x)) = f^-1(f(x)) then their derivatives have the relationship that:\[(f^{-1})'=\frac 1{f'}\]

Answer 7

forgot the =x in that lol

if ..f(f^-1(x)) = f^-1(f(x)) = x

Answer 8

in this case:

what is the derivative of: sqrt(3x+1) ?

Answer 9

Good job, Amistre