IF g(x) is inverse of f(x) and f ' (x) = 1/1+x^2 then g' (x) =

IF g(x) is inverse of f(x) and f ' (x) = 1/1+x^2 then g' (x) =?
answer is:::: 1+(g(x))^2
please explain it in easiest way because i am in learning stage.
answers of multiple peoples will be appreciated.

Take this problem step by step.

You need to know what g'(x) is.
To get this, you need to know what g(x) is.
To get g(x), you need to know f(x).
Thankfully, you can work out f(x).

f(x) = integral f'(x) dx = integral [ 1/(1+x^2) ] dx

Standard integral from your integral tables, the solution is:

f(x) = tan^-1(x)

Since g(x) is the INVERSE of f(x), it follows:

g(x) = tan(x)

Taking the derrivative:

g'(x) = sec^2(x)

Consider g(x) = tan(x)

>> g(x)^2 = tan^2(x)
>> g(x)^2 = sec^2(x) - 1
>> g(x)^2 + 1 = sec^2(x)
>> g(x)^2 + 1 = g'(x)

Re-arrange:

>> g'(x) = 1 + (g(x))^2

Which was to be shown.

g(x) is inverse of f(x) -----> f(g(x)) = x

Now differentiate both sides with respect to x:
f'(g(x)) * g'(x) = 1
g'(x) = 1 / f'(g(x))
g'(x) = 1 / [1 / (1 + (g(x))²)
g'(x) = 1 + (g(x))²