Chain Rule: The General Logarithm Rule - Problem 2
We’re differentiating functions that are compositions of natural log and some other function. And in the next example, I’m going to do a problem where a property of natural logs will come in very handy. This property; natural log of A/B equals natural log of A minus natural log of B. Now you don’t need this property but its going to make the problem a lot easier to solve. So let’s take a look.
It says differentiate h(x) equals ln of 9 minus x over 2x plus 3. So I’m going to use this property of logs I just introduced to break this function up into 2 pieces. The log of a quotient can be broken up into the log of the numerator minus the log of the denominator. That’s just from properties of logs. Now when I differentiate it I can differentiate it in two pieces.
Now the derivative of the log of a function is 1 over that function. 1 over 9 minus x times the derivative of the function. The derivative of 9 minus x, that’s -1, minus, and the derivative of this guy is 1 over 2x plus 3 times the derivative of 2x plus 3, which is 2. So just to simplify, I get -1 over 9 minus x, minus 2 over 2x plus 3.
Now whenever you have -1 over 9 minus x, something like this, you can multiply any fraction top and bottom by -1. It doesn’t change the value of the fraction because you are effectively multiplying by 1. But let’s say we awn tot get a positive numerator for some reason or we’d like to have the x in front in the denominator. We multiply the top and bottom by -1 and we get +1 on top. -9 plus x is the same as x minus 9. That’s a much nicer looking answer. That’s my answer; the derivative of ln of this complicated quotient is 1 over x minus 9 minus 2 over 2x plus 3.