Derivative of 1 − 5 − x Any help is greatly appreciated!I have tried using logs, but I don't think it is correct; y = 1 − 5 − x ln ⁡ ( y ) = ln ⁡ ( 1 ) + x ln ⁡ ( 5 ) y = x ln ⁡ ( 5 )and hence y ′ = x 5

Question

Derivative of   1  −      5          −      x      Any help is greatly appreciated!I have tried using logs, but I don&#039;t think it is correct;  y  =  1  −      5          −      x        ln  ⁡  (  y  )  =  ln  ⁡  (  1  )  +  x  ln  ⁡  (  5  )  y  =  x  ln  ⁡  (  5  )and hence      y    ′    =      x    5

jockedripperHZT · Accepted Answer

You can solve the problem using the idea of logs but note that the equality you wrote is wrong. You have  f  (  x  )  =  1  −      5          −      x        =  1  −      e          ln      ⁡      (              5                  −          x                    )        =  1  −      e          −      x      ln      ⁡              5            and so using the chain rule you have      f    ′    (  x  )  =  −      e          −      x      ln      ⁡              5              ⋅            (      −      x      ln      ⁡              5            )        ′    =  −      e          −      x      ln      ⁡              5              (  −  ln  ⁡      5    )  =      e          −      x      ln      ⁡              5              ln  ⁡      5    =      5          −      x        ln  ⁡      5    .

Derivative of 1-5^(-x)

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