A function and its inverse walked into a bar!

$\Large f(3)=4 \Rightarrow g(4)=3$ For inverse functions, $\LARGE f(g(x))=x$

Differentiating both sides w.r.t x we get, $f'(g(x))•g'(x)=1 \Rightarrow g'(x)=\frac{1}{f'(g(x))}$ Putting $x=4$ $g'(4)=\frac{1}{f'(g(4))} \Rightarrow g'(4)=\frac{1}{f'(3)}=(5/2)^{1/3}$

Differentiating again, $\Large f'(g(x))•g''(x)+f''(g(x))•g'(x)•g'(x)=0$ Putting again $x=4$ we finally get, $g''(4)=(\frac{5}{2})^{\frac{1}{3}+\frac{2}{3}}•(\frac{4}{7})=\frac{10}{7}$ Answer $10+7=\boxed{17}$

$\color{#3D99F6}{\textbf {A very nice question BTW}}$

Since $f$ and $g$ are inverses, we know that $g(f(x)) = x$ .

Differentiating both sides with respect to $x$ using the chain rule, we have that

$g'(f(x)) * f'(x) = 1 \Longrightarrow g'(f(x)) = \dfrac{1}{f'(x)}$ .

Next, differentiate $g'(f(x)) * f'(x) = 1$ using the product and chain rules to find that

$g''(f(x)) * (f'(x))^{2} + g'(f(x)) * f''(x) = 0$

$\Longrightarrow g''(f(x)) = -\dfrac{g'(f(x)) * f''(x)}{(f'(x))^{2}} = -\dfrac{f''(x)}{(f'(x))^{3}}$ ,

since $g'(f(x)) = \dfrac{1}{f'(x)}$ .

Thus $g''(4) = g''(f(3)) = -\dfrac{f''(3)}{(f'(3))^{3}} = \dfrac{\frac{4}{7}}{\frac{2}{5}} = \dfrac{10}{7}$ .

Thus $a = 10, b = 7$ and $a + b = \boxed{17}$ .

Since we are looking for the second derivative of g, we can find the derivative of the first derivative of g. The first derivative of g is just the inverse rule:

$g'\left( x \right) =\frac { 1 }{ f'(g(x)) }$

We differentiate both sides:

$\frac { d }{ dx } g'\left( x \right) =\frac { d }{ dx } (\frac { 1 }{ f'(g(x)) } )$

Using the quotient rule:

$\frac { d }{ dx } (\frac { 1 }{ f'(g(x)) } )=\frac { f'(g(x))(\frac { d }{ dx } 1)-(1)(\frac { d }{ dx } f'(g(x))) }{ { [f'(g(x))] }^{ 2 } }$

Since the derivative of 1 is 0, the first part of the numerator disappears

$\frac { -\frac { d }{ dx } f'(g(x)) }{ { [f'(g(x))] }^{ 2 } }$

To finish simplifying, we use the chain rule:

The derivative of the outside function $f'(x)$ is $f''(x)$ The derivative of the inside function $g(x)$ is just $g'(x)$ We can also consider $-1$ to be a constant that we can factor out to make it easier for us

$=-\left[ \frac { f''(g(x))(g'(x)) }{ { [f'(g(x))] }^{ 2 } } \right]$

However, $g'(x)$ is can be rewritten as $\left\{ \frac { 1 }{ f'(g(x)) } \right\}$ due to the inverse rule that we established at the beginning.

$-\left[ \frac { f''(g(x))\left\{ \frac { 1 }{ f'(g(x) } \right\} }{ { [f'(g(x))] }^{ 2 } } \right] =-\left[ \frac { f''(g(x)) }{ { [f'(g(x))] }^{ 3 } } \right]$

Now we can plug in numbers. $g(4) = 3$ so we have $-\left[ \frac { f''(3) }{ { [f'(3)] }^{ 3 } } \right]$

$= -\left[ \frac { -\frac { 4 }{ 7 } }{ { [\sqrt [ 3 ]{ \frac { 2 }{ 5 } } ] }^{ 3 } } \right] =\frac { \frac { 4 }{ 7 } }{ \frac { 2 }{ 5 } } =\frac { 4 }{ 7 } \times \frac { 5 }{ 2 } =\frac { 10 }{ 7 }$

$10 + 7 = 17$