Question

Suppose f(x)= sin (pi cos x) .On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with respect to x is:

Answer 1

$$
\Large f(x) = \sin (\pi \cos(x))
$$

Answer 2

is that the correct f(x)

Answer 3

yes

Answer 4

$$
\Large {
y = \sin (\pi \cos( x) )
\\ \therefore \\
x = \sin (\pi \cos (y) )
\\ \implies \\
\arcsin(x) = \pi \cos( y)
\\ \cos y = \frac{1}{\pi} \arcsin(x)
\\ \implies\\

\\ y = \arccos (\frac{1}{\pi} \arcsin(x))
\\ \implies\\

\frac{dy}{dx}=- \frac{1}{\sqrt{1-(\frac{1}{\pi} \arcsin(x))^2~})}\cdot ...
}

$$

Answer 5

is this a multiple choice question

Answer 6

The answers are:

-1/cos(pi cosx) where x and y are related by the equation (satisfy the equation) x= sin(pi cosy)

-1/ pi sin x cos (pi cos x) where x and y are related by the equation x= sin(pi cosy)

-1/ pi sin y cos (pi cos y) where x and y are related by the equation x= sin(pi cosy)

-1/ cos (pi cos y) where x and y are related by the equation x= sin(pi cosy)

-1/sin y cos (pi cos y) where x and y are related by the equation x= sin(pi cosy)

Answer 7

@perl

Answer 8

there is a trick we can use

Answer 9

$$
f(x) = \sin (\pi \cos(x))\\

\LARGE \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}
\\ \Large {
\\ \therefore \\
\frac{dx}{dy} = \frac{1}{\cos (\pi \cos(x))\cdot \pi \cdot (-\sin(x)} \\


}
$$

Answer 10

So then it would be the second one.

Answer 11

we have to be careful, this is the inverse with respect to y

Answer 12

So then?

Answer 13

i see your choices, some are in terms of y, some in terms of x

Answer 14

Is it x or y?

Answer 15

I believe y

Answer 16

you can test this supposition with
y = x^3
you know the inverse is
y = x^(1/3)
and the derivatives of these

Answer 17

$$ \Large Given ~~
f(x) = \sin (\pi \cos(x))\\

\LARGE \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}
\\ \Large {
\\ \therefore \\
\frac{dx}{dy} = \frac{1}{\cos (\pi \cos(x(y)))\cdot \pi \cdot (-\sin(x(y)))} \\
\\ \therefore \\ ~\\
(f^{-1}(x) )' = \frac{1}{\cos (\pi \cos(y(x)))\cdot \pi \cdot (-\sin(y(x)))} \\
(f^{-1}(x) )' = \frac{1}{\cos (\pi \cos(y))\cdot \pi \cdot (-\sin(y))} \\

}

$$

Answer 18

Alright thanks!