Step-by-step explanation:

For the SAKE of easier notation, we shall say that

f

is a function with inverse function

g

, that is,

f

−

1

(

x

)

=

g

(

x

)

.

According to the definition of inverse functions,

f

(

g

(

x

)

)

=

x

Differentiating through the chain rule gives

f

'

(

g

(

x

)

)

g

'

(

x

)

=

1

Solving for the derivative of the inverse gives

g

'

(

x

)

=

1

f

'

(

g

(

x

)

)

So, we want to find

g

'

(

2

)

.

g

'

(

2

)

=

1

f

'

(

g

(

2

)

)

We want to first find

g

(

2

)

, however, we cannot write an expression for

g

(

x

)

. What we have to remember is that the domain of the MOTHER function is the range of its inverse function, and vice versa.

Note that if

f

(

x

)

=

2

, then

x

=

g

(

2

)

. This means we should let

f

(

x

)

=

2

then solve for

x

, which is equal to

g

(

2

)

.

f

(

x

)

=

2

⇒

x

5

+

3

x

−

2

=

2

Continuing to solve yields

x

5

+

3

x

−

4

=

0

This may look impossible, but notice that the SUM of the coefficients of each term is

0

, that is,

1

+

3

−

4

=

0

. This means that

x

=

1

is a solution.

Dividing, we see that

(

x

−

1

)

(

x

4

+

x

3

+

x

2

+

x

+

4

)

=

0

Note that

x

4

+

x

3

+

x

2

+

x

+

4

>

0

for all values of

x

, so it has no REAL roots. Thus,

f

(

x

)

=

2

when

x

=

1

, i.e.,

f

(

1

)

=

2

.

This means that

g

(

2

)

=

1

.

Returning to what we had earlier, we see that

g

'

(

2

)

=

1

f

'

(

g

(

2

)

)

=

1

f

'

(

1

)

Find this by taking the derivative of

f

.

f

(

x

)

=

x

5

+

3

x

−

2

⇒

f

'

(

x

)

=

5

x

4

+

3

Thus

f

'

(

1

)

=

5

+

3

=

8

.

g

'

(

2

)

=

1

8

With your notation,

[

f

−

1

]

'

(

2

)

=

1

8

my friend, hope it will help u