A calculus problem get points!?

Question

If f(x)=y=sin(x)+cos(x), and g is the inverse of y, then what is dg/dx at x=-1 (derivative of g at x=-1).

Show me how you worked it out, not just the solution. I think you have to use the formula dg/dx=1/(g(dy/dx(-1))). I do not know how to work out the inverse of y, since I don't know how to express x in terms of y.

Please help!

Answer 1

y=cosx+sinx
y=cosx+(1-cos^2x)^(1/2)
square on both sides.
1-cos^2x+cos^2x+2cosx(1-cos^2X)^(1/2)=y^2
cos^2 x(1-cos ^2 x)=(y^2-1)/2
sin 2x = y^2-1
x = (1/2) Sin^(-1) (y^2-1)
f^(-1) (X) = (1/2) Sin^(-1) (x^2-1)=g
dg/dx = .5 x (d/dx) Sin^(-1) (x^2-1)
I don't even know. I'm in my 4th year of Engineering. I've forgotten all that shi*t.

Answer 2

df/dx= cos x-sin x .You don´t need to work out the inverse function
dx/df =1/(cosx-sin x)

x=g(y) at y = -1 which means sin x+ cos x = -1
squaring 2sin x cos x=0
sin 2x = 0 and 2x =+2kpi
and x=kpi zero is not sopution
At x= pi sin x + cos x=-1
As the derivative of the inverse function is the inverse of the derivative of the direct function at the corresponding point
this is in the given problem x= pi
and dx/dy=1/cospi-sin pi) =-1)
The confusion arises because as the inverse function is x=g(y) when you then incherchanges x and y the value x=-1 really means y=-1