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

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

stops on an even number.