Question

You are given the functions f(x) and g(x) = sin^-1(fx). Which of the following expressions represents the ratio of the derivatives of these two functions, f'(x)/g'(x)? Assume that the domain of g(x) is -pi/2 < f(x) < pi/2.

A. f'(x)/g'(x) = 1/(sqrt( 1 - (f(x))^2 )
B. f'(x)/g'(x) = 1/(sqrt( 1 + f(x) )
C. f'(x)/g'(x) = (sqrt( 1 - (f(x))^2 )
D. f'(x)/g'(x) = (sqrt( 1 - f(x) )
E. f'(x)/g'(x) = (sqrt( 1 + f^2(x) )

Answer 1

This isn't the answer as you can see it above, but do you have the rest of the questions?

Explanation:

Thank you in advance.

Answer 2

C

Explanation:

We are given the two functions:

`\displaystyle f(x) \text{ and } g(x)=\sin^(-1)(f(x))`

And we want to find the ratio that represents:

`\displaystyle (f^\prime(x))/(g^\prime(x))`

To do so, we will compute the derivatives of both functions.

For f, we can simply write that:

`f^\prime(x)=f^\prime(x)`

For g, we will use the chain rule:

`\displaystyle (u(v(x))^\prime=u^\prime(v(x))\cdot v'(x)`

We can let:

`u(x)=\sin^(-1)(x)\text{ and } v(x)=f(x)`

We can double check this by doing the composition:

`g(x)=u(v(x))=\sin^(-1)(f(x))`

Therefore, by the chain rule, we acquire that:

`\displaystyle g^\prime(x)=(1)/(√(1-(v(x))^2))\cdot v'(x)`

By substitution:

`\displaystyle g^\prime(x)=(f^\prime(x))/(√(1-(f(x))^2))`

Hence, our ratio is:

`\displaystyle (f^\prime(x))/(g^\prime(x))=(f^\prime(x))/((f^\prime(x))/(√(1-(f(x))^2)))`

Simplify:

`\displaystyle (f^\prime(x))/(g^\prime(x))=√(1-(f(x))^2)`

Hence, our answer is C