# If F(x) is a differentiable function, and F(-1) = 5, F'(-1) = 4 and (F^-1)(4) = 2/3 then what is (F^-1)'(5)?

Jan 22, 2017

$\left({F}^{-} 1\right) ' \left(5\right) = \frac{1}{4}$.

#### Explanation:

$\left({F}^{-} 1 \circ F\right) \left(x\right) = x \Rightarrow {F}^{-} 1 \left(F \left(x\right)\right) = x$

Diff.ing both the sides w.r.t $x$ using the Chain Rule, then,

(F^-1)'((F(x))F'(x)=1.

$\text{Taking, "x=-1," we get, } \left({F}^{-} 1\right) ' \left(F \left(- 1\right)\right) F ' \left(- 1\right) = 1$.

Substituting the given values, we get,

$\left\{\left({F}^{-} 1\right) ' \left(5\right)\right\} \left(4\right) = 1$

$\therefore \left({F}^{-} 1\right) ' \left(5\right) = \frac{1}{4}$

Taking the other way round, $\left(F \circ {F}^{-} 1\right) \left(x\right) = x$

$\therefore F \left({F}^{-} 1 \left(x\right)\right) = x \Rightarrow \frac{d}{\mathrm{dx}} \left\{F \left({F}^{-} 1 \left(x\right)\right)\right\} = \frac{d}{\mathrm{dx}} \left(x\right)$

$\Rightarrow F ' \left({F}^{-} 1 \left(x\right)\right) \left\{{F}^{-} 1 \left(x\right)\right\} ' = 1$

"Taking "x=5, F'(F^-1(5){F^-1(5)}'=1...............(star)

Recall that $F \left(- 1\right) = 5 \Rightarrow {F}^{-} 1 \left(5\right) = - 1$

$\therefore F ' \left({F}^{-} 1 \left(5\right)\right) = F ' \left(- 1\right) = 4 , \text{ and, so, by } \left(\star\right) ,$ we have,

$4 \left\{{F}^{-} 1 \left(5\right)\right\} ' = 1 \therefore \left({F}^{-} 1\right) ' \left(5\right) = \frac{1}{4}$.