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

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

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answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

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