Calculus Derivatives and Integrals of lnX
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Derivatives and Integrals of lnX


 The twin of exponential equations are logarithms.

 If we say ...

 Y = eX

 That also means ...

 log eY = X

 (also known as ln Y = X)

 Remember the way you read a log is:


 Which is the same thing we said in the beginning ...

 Y = eX


 (NOTE: From here on we'll be using ln instead of loge)

 So the point is:

 Y = eX    and   ln Y = X

 Say the exact same thing.
 So let's talk about the derivatives and integrals of things like

 ln X

 When we went looking for the derivative of eX, we found that it was eX,
 give or take a dX.
 Wouldn't it be nice if ln X worked the same way
 SORRY, No such luck.
 But finding it isn't too bad.
 Start with 

 Y = ln X

 which is the same thing as saying:

X = eY 

 (Don't worry, we'll be back to ln X before we're done)

 OK, the derivative of eY (with respect to the variable Y) is eY
 so we can say:


 We can flip these over and say:


 Now we do some substitution.
 We started with the equations:


  Y = ln X   also known as   X = eY 

 Now we substitute both of these into our last equation:


 In English, what this says is that the derivative of lnX,
 is equal to 1/X (also known as X-1)
 This also means that the integral of X-1 is lnX,
 remembering, of course, to include dX's and + C's where needed.


 Sometimes you will see written as
 Don't let that throw you, they both mean the same thing.


 Did you notice the 2X in the numerator of the last one?
 The derivative of X2 is 2XdX,  so everything we needed to have,
 to "Put Back Into" the 1/X2 was there.
 So this all works out great for things like eX and ln X,
 but what if you have to deal with 2X or log8X ?

   copyright 2005 Bruce Kirkpatrick

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