# Given f=e^x+x+15 what is (f^-1)'(1).

### 3 Answers | Add Yours

f= e^x + x + 1

First let us find the derivative:

f'(x) = e^x + 1

Now we know that:

f'(x) * (f^-1)' = -1

Now substitute:

(e^x+1) *(f^-1)' = -1

==> f^-1)' = -1/(e^x + 1)

Now to find (f^-1)'(1) we will substitute with x= -1:

==> (f^-1)(1) = -1/(e^1 + 1)

= -1/(e+1)

** ==> (f^-1)(1) = -1/(e+1)**

f(x) = e^x+x+15.

f(x) is a continuous monotonosly increasing function as e^x and x+15 are also continous monotonously incresing. So f(x) = e^x+x+15 is a bijection . So the f^-1 function exists.

To find f^(-1)'(1).

f(x) = e^x+x+15,

Let y = f^(-1)x.

Then f(y) = x.

x = e^y+y+15 by definition

Differentiate with respect y.

dx/dy = e^y+1.

Take the reciprocal .

dy/dx = 1/(e^y+1)

(f^-1)' (x) = 1/e^y +1

(f^'-1) (x)= 1{ e^(e^x+x+15) +1}

(f^-1)'(-1) = 1/{e^(e+1+15) +1}

(f^-1)(1) = 1/{e^(e+16) +1}.

Since the requestof the problem is to calculate the derivative of the inverse of f(x), we'll conclude that the given function is bijective.

According to the rule, a bijective function is invertible.

We know that the product between the derivative of a function and the derivative of the inverse of the function is -1.

So, we'll write mathematically the statement:

[f(x)]'*[(f(x))^-1]' = 1

We infer that (f^-1)' = 1/[f(x)]'

We'll differentiate the function f(x):

f'(x) = (e^x + x + 15)'

f'(x) = e^x + 1

So,

(f^-1)' (x) = 1 / (e^x + 1)

Now, we'll calculate (f^-1)' (1):

(f^-1)' (1) = 1 / (e^1 + 1)

**(f^-1)' (1) = 1/(e+1)**

**(f^-1)' (1) = 0.264550...**