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well the question is find g'(-1) where g is the inverse of f
i don't think that form is of any use to me
g'(x) = 1 / (f'(g(x))
take the derivative and solve for dy/dx
sob, i wish i had math vision like you
rusty here ...
who are you anyway, teacher?
not done yet ...
find y in the main equation when x = -1
did you get the right answer?
did you ask me if I'm a teacher ?
as long as you keep the cross product the same the expressions are quivalent
i never seen just "dx" alone used like that before, didn't know you could do that
dx is the derivative of x
I know this method is easier than the one you have ...
have you ever read Calculus made easy by Thompson ?
i'm confused, why did we take the derivative of that again?
read it ! you'll see methods that are hidden from students !
Ok ...where are you confused ?
I just did it once !
I'll write the steps down
f(x) = x^5 + 2x^3 + x - 1, find the inverse derivative (g'(b)) b = -1
the answer is 1 says my textbook
yes .... x is y in the inverse ... darn
if g is the inverse of f, then g' = dy/dx = 1/5y^4 + 6y^2 + 1 right?
then g'(-1) = 1 / 5(-1)^4 + 6(-1)^2 + 1?
what is the answer in your book?
yeah i don't know, this question isn't even important for class
hold on ....it shouldn't be this hard !
i'll type out the entire question, maybe i asked it wrong..
show point (a,b) lies on graph of f f(x) = x^5 + 2x^3 + x - 1 ; (0,-1) find g'(b) where g = inverse of f
don't worry about it, it's a dumb question and not even gonna be on the exam
i'm about to pass out x_x
oh, and yeah, are you a teacher or something? masters in math or something
you know ...I'll solve it and send it to you ...
with both methods
hahah that'd be nice
should go help someone else though, reallocate your skills
gotta sleep, i'll see you around mathg8
the slope of the inverse function at (c,d) is the reciprocal of the slope of the original function at (d,c)