Calculus Question?

Question

Let f(x) = x3 + x. If h is the inverse function of f, then h'(2) =

(A) 1/13
(B) 1/4
(C) 1
(D) 4
(E) 13

I know the correct answer is b but i have no idea how they got it. Can you please explain how they got that answer

Answer 1

There's a formula for finding the derivative of the inverse function at a particular point WITHOUT actually knowing what the inverse function is:

(Writing f^(-1) as the inverse function of f)

[f^(-1)]'(a) = 1 / (f'(f^(-1)(a)).

Sorry if that's hard to read-- in English it says, to find the derivative of the inverse of f at a point a, find the inverse of a, evaluate the derivative of f at that point, and then take 1 divided by that number.

In your problem, (now using h for f inverse), h(2) = x where x^3 + x = 2. Normally you'd have to do some factoring to solve this, but you can see that h(2) = 1 here, because 1^3 + 1 = 1 + 1 = 2.

Now, f'(x) is 3x^2 + 1, so f'(1) = 3(1)^2 + 1 = 4.

As the last step, take 1 divided by this number -> 1/4.

Another way to solve this problem is to actually find f inverse (or h in this problem), take its derivative, and evaluate at 2, but that's a huge pain... much simpler to use the above formula.

Answer 2

I could say structure is a facet of engineering this means that that a well seize on math is fundamental, pre-cal is is like algebra to the following degree with the Pi chart and figuring out approximately radians and attitude measures (which turns out primary) however Calculus is the research of derivatives, (the slope of a line at a unique factor) sort of needless it doesn't matter what subject of workd your going to, in case you question me.

Answer 3

dy/dx = 3x^2+1 so dx/dy = 1/(3x^2+1) where I wrote x( instead of y as it is the inverse functions derivative)
The x corresponding to y = 2 is
2=x^3+x which yields x=1
so dx/dy=1/(3+1) =1/4 (B)

Answer 4

suppose x = h(y)
y = h^3 + h
1 = 3h^2 h' + h'
at y = 2 I can see x = 1 = h(2)
1 = 4h'
so h'(2) = 1/4

Answer 5

You should see
http://www.analyzemath.com/calculus/Differentiation/derivative_inverse.html