Let f(x)=10x+(arctanx)^2+2. If g(x) = f-1(x)

Question

Let f(x)=10x+(arctanx)^2+2. If g(x) = f-1(x) <- inverse. If g is the inverse function of f, then find the value of g'(2)

so far I have got the derivative of g(x)
f'(y) = 10y'+2(arctany)y'/1+y^2
and I can't figure out what to do with the y^2

dumbcow · Answer

$$g'(2) = \frac{1}{f'(g(2))}$$ $$f'(x) = 10 + \frac{2 \tan^{-1} x}{1+x^{2}}$$ g(2) is the solution to f(x) = 2 $$10x + (\tan^{-1} x)^{2} +2 = 2$$ $$10x +(\tan^{-1} x)^{2} = 0$$ x = 0 http://www.wolframalpha.com/input/?i=10x+%2B+arctan%28x%29%5E2+%3D+0 $$f'(0) = 10 + 2\tan^{-1} 0 / 1+0= 10$$ $$g'(2) = \frac{1}{10}$$

anonymous · Answer

thank you so much!

dumbcow · Answer

no problem... sorry i kinda got lost trying to follow your work, i like to just keep everything in terms of x

anonymous · Answer

It was my fault should've made it more clearer :(

dumbcow · Answer

you took the derivative correctly though just looks like you used implicit differentiation ?

anonymous · Answer

yes I did implicit differentiation, but that just got me more confused. 
Your method was much easier to understand the question :)

dumbcow · Answer

:)