# let `f(x)=x^19+x^17+x` find `(f^-1)'(3)`

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You need to remember the following equation that relates the derivative and its inverse, such that:

`f'(3)*(f^(-1)(3))' = 1 => (f^(-1)(3))' = 1/f'(3)`

You need to evaluate the derivative of the function, such that:

`f'(x) = 19x^18 + 17x^16 => (f^(-1)(x))' = 1/19y^18 + 17y^16`

Evaluating the derivative of the function at `x = 3` yields:

`f'(3) = 19*3^18 + 17*3^16`

Factoring out `3^16` yields:

`f'(3) = 3^16*(19*9 + 17) => f'(3) = 3^16*188`

Replacing `3^16*188` for `f'(3)` yields:

`(f^(-1)(3))' = 1/(3^16*188)`

**Hence, evaluating the derivative of inverse function, at `x = 3, ` yields **`(f^(-1)(3))' = 1/(3^16*188).`