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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) )


Sagot :

Answer:

C

Step-by-step explanation:

We are given the two functions:

[tex]\displaystyle f(x) \text{ and } g(x)=\sin^{-1}(f(x))[/tex]

And we want to find the ratio that represents:

[tex]\displaystyle \frac{f^\prime(x)}{g^\prime(x)}[/tex]

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

For f,  we can simply write that:

[tex]f^\prime(x)=f^\prime(x)[/tex]

For g, we will use the chain rule:

[tex]\displaystyle (u(v(x))^\prime=u^\prime(v(x))\cdot v'(x)[/tex]

We can let:

[tex]u(x)=\sin^{-1}(x)\text{ and } v(x)=f(x)[/tex]

We can double check this by doing the composition:

[tex]g(x)=u(v(x))=\sin^{-1}(f(x))[/tex]

Therefore, by the chain rule, we acquire that:

[tex]\displaystyle g^\prime(x)=\frac{1}{\sqrt{1-(v(x))^2}}\cdot v'(x)[/tex]

By substitution:

[tex]\displaystyle g^\prime(x)=\frac{f^\prime(x)}{\sqrt{1-(f(x))^2}}[/tex]

Hence, our ratio is:

[tex]\displaystyle \frac{f^\prime(x)}{g^\prime(x)}=\frac{f^\prime(x)}{\dfrac{f^\prime(x)}{\sqrt{1-(f(x))^2}}}[/tex]

Simplify:

[tex]\displaystyle \frac{f^\prime(x)}{g^\prime(x)}=\sqrt{1-(f(x))^2}[/tex]

Hence, our answer is C