# Thread: How would you solve this problem?

1. On my Calc II quiz today, I couldn't figure out how to do this one problem, and it's driving me nuts just thinking about it... How would you solve it?

f(x)=x^3+3sin(x)+2cos(x)
a=2

Find f-1'(a)

Note that that is the derivative of the inverse function of f. This problem would be easy if I could solve for f-1 and plug it into f-1'(x)=1/(f'(f-1(x)), but I couldn't see any way to do that. I racked my brain thinking of a way to implicitly solve for it, but I came up with a blank. Any help?  2.

3. I'm not sure i've understood from the question how 'a' is related to f(x), i presume that in which case the derivative of the inverse of f should be 1 for all values of 'x'... shouldn't it?

Did the quiz explicitly state that you must find the derivative of the inverse? If you hadn't specified it i would have taken this to be the inverse of the derivative.  4. Somewhere in your text or studies you should have seen a simple relationship between the first derivative of a function and the first derivative of its inverse function, at a point.

The mental picture is usually the slopes of the two tangent lines specified, with the words "negative" and "reciprocal". The one function is sometimes thought of as "stretching" the domain, which must then be "contracted" to get back in the inversion (or vice versa). Occasionally people imagine flipping the graph around, swapping the axes, etc.  calculus, derivative, inverse 