If f(x) = arcsinx, then f' (√3/2)= How do I find the derivative of the inverse in this problem?

You can either use the inverse function theorem or compute the general derivative using implicit differentiation. The first method is slightly faster.

The IFT goes like this: if f(x) is invertible and f(a) = b, then finv(b) = a (where "finv" means "inverse of f").

By definition of inverse functions, we have

f(finv(x)) = finv(f(x)) = x

Differentiating both sides of the second equality with respect to x using the chain rule gives

finv'(f(x)) * f'(x) = 1

When x = a, we get

finv'(b) * f'(a) = 1

or

finv'(b) = 1/f'(a)

Now let f(x) = sin(x), which is invertible over the interval -π/2 ≤ x ≤ π/2. In the interval, we have sin(x) = √3/2 when x = π/3. We also have f'(x) = cos(x).

So we take a = π/3 and b = √3/2. Then

arcsin'(√3/2) = 1/cos(π/3) = 1/(1/2) = 2

for finding the midpoint of any given 2 pair of points simply add the respective x coordinates with other x coordinate and the y coordinate with other y coordinate and divide them by 2

in your question it will be :

(3+1)/2 for the x coordinate of mid point i.e 2

[8+(-4)]/2 for the y coordinate of midpoint i.e 2

so coordinates of mid point of points (3,8) and (1,-4) will be (2,2)

what is the question here

step-by-step explanation: