Derivative of a Inverse Function: If g(x) = x^5 + x^{27}, what is (g^{-1})'(2)?

[Baris]

I used (f^-1)'(a) = 1/f'(f^-1(a)) but the answer always turns out to be wrong.

If $g(x) = x^5 + x^{27}$, what is $\left(g^{-1}\right)'(2)$? (Hint: $g(1) = 2$)

Can anybody help?

 

[khansaheb]

I used (f^-1)'(a) = 1/f'(f^-1(a)) but the answer always turns out to be wrong. Can anybody help?
View attachment 36586

What are the unknowns (variables) of this problem?

Now from the given information in the problem, set up the equation.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

If I were to work on this problem, I would start with:

f[f^-1] = x ..... and use chain-rule of differentiation.

 

[stapel]

I used (f^-1)'(a) = 1/f'(f^-1(a)) but the answer always turns out to be wrong.

If $g(x) = x^5 + x^{27}$, what is $\left(g^{-1}\right)'(2)$? (Hint: $g(1) = 2$)

Please reply *showing* what you did after applying the referenced formula.

 

[Steven G]

(g^-1(y))' = 1/g'(x).
You want (g^-1(2))' which equals 1/g'( ?????).
When y =2, then x=????. Just look at the hint.
Continue.....

 

[Steven G]

I used (f^-1)'(a) = 1/f'(f^-1(a)) but the answer always turns out to be wrong.

Yes, that is correct but it is cleaner to replace f^-1(a) with b, where f(b)=a