Question
Use the function f and the given real number a to find (f^-1)'(a)
f(x)=(x+9)/(x+3), x>-3, a=2
f(^-1)'(a)= ?
Can someone please show me how to solve this problem?
f(x)=(x+9)/(x+3), x>-3, a=2
f(^-1)'(a)= ?
Can someone please show me how to solve this problem?
Answers
Damon
oh, you want the derivative of the inverse function
x = (y+9)/(y+3)
dx = [(y+3)dy - (y+9)dy]/(y+3)^2
dx = -6 dy/(y+3)^2
dy/dx = (y+3)^2/-6
x = (y+9)/(y+3)
dx = [(y+3)dy - (y+9)dy]/(y+3)^2
dx = -6 dy/(y+3)^2
dy/dx = (y+3)^2/-6