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anonymous

  • one year ago

If f(x) = x+ tan x and g(x) is the inverse of f(x) then g'(x) is equal to..

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    aren't you trying to find the derivative of inverse at a particular point ?

  3. anonymous
    • one year ago
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    answer is 1/(2+{g(x) - x}^2) and i have no clue

  4. Empty
    • one year ago
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    That doesn't seem right to me

  5. anonymous
    • one year ago
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    at a particular point it would be easy na

  6. ganeshie8
    • one year ago
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    maybe start with this : \[f(g(x)) = x\]

  7. Empty
    • one year ago
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    Since it's the inverse I plugged it in to get: \[x=g(x)+\tan g(x)\] Then differentiate: \[1=g'(x)+g'(x)\sec^2g(x) \] \[g'(x)=\frac{1}{1+\sec^2 g(x)}\]

  8. Empty
    • one year ago
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    Well that's why I said it doesn't seem right idk depends on what kind of answer you're looking for I guess.

  9. anonymous
    • one year ago
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    thanks i get it... converting sec^2 g(x) to tan ^2 g(x) then substituting tan^2 from the eqn f(g(x)) = x

  10. ganeshie8
    • one year ago
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    Nice :)

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