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# Differentiating logarithmic functions using log properties

## Video transcript

let's say that we've got the function f of X and it is equal to the natural log of X plus 5 over X minus 1 and what we want to figure out is what is f prime of X and I encourage you to pause this video and try to figure it out on your own so there's two ways that you could approach this I would call one way the easy way and the other way the hard way and we'll work through both of them the easy way is to recognize your logarithm properties to remember that the natural log of a over B remember natural log is just log base the number E so this is just going to be equal to the natural log of a minus the natural log of B so if we just apply this property right over here to simplify this expression or at least simplify it from a point of view in terms of having to take its derivative we can rewrite f of X we can write f of X as being equal to the natural log of x plus 5 minus the natural log of X minus 1 and when we take the derivative now with respect to X F prime of X well this is going to be the derivative of the natural log of X plus 5 with respect to X plus 5 so that's going to be 1 over X plus 5 times the derivative of X plus 5 with respect to X I'm just applying the chain rule here and that's just going to be 1 so this the derivative of that is that and the derivative of this well let's see we're going to have a minus sign there and the derivative of natural log of X minus 1 with respect to X minus 1 is going to be 1 over X minus 1 and then the derivative of x minus 1 with respect to X is just 1 so you just multiply this by 1 doesn't really change the value and we're done we were able to figure out the derivative of F now what's the hard way you might be thinking or maybe you did do it when you try to approach it on your own well that's not to simplify this expression using this property and just to try to power through this using the chain rule so let's try to do that so in that case f prime of x it's going to be the derivative of this whole thing with respect to X plus five over X minus one which is going to be 1 over X plus 5 over X minus 1 times the derivative times the derivative with respect to X of X plus 5 over X minus 1 this is just the chain rule derivative of this whole thing with respect to this expression times the derivative of this expression with respect to X just the chain rule so let's see this is going to be equal to let me use some colors here what this one I'm boxing off in blue that's the same thing as X minus 1 over X plus 5 I'm just taking the reciprocal of this and then it's going to be x and then I'll do this in magenta that's not magenta so that's going to be x and I'm going to rewrite it as the derivative with respect to X of X plus 5 times X minus 1 to the negative 1 power and I like to write it that way because I always forget I always forget the whole quotient rule thing but I remember the product rule so this thing so let me just rewrite it then I think you already appreciate why this is the hard way so let me write this this is X minus 1 over X plus 5 times so let's apply the product rule derivative of x plus 5 well that's just 1 times the second term times X minus 1 to the negative 1 so that's 1 over X minus 1 and then plus what's the derivative of X minus 1 to the negative 1 well let's see that's going to be that's going to be negative X minus 1 to the negative 2 power so I could say negative 1 X minus 1 to the negative 2 or I could just write it like this and then times the derivative of X minus 1 with respect to X well that's just going to be 1 and then times this X plus 5 so actually so x times X plus five so all I did here product rule derivative derivative of this is one times that and that gave us that right over there and then I took the derivative of this which is this right over here negative one over X minus one of the over X minus one squared or you could say negative X minus one of the negative two power times this first expression right over there so that's that derivative and now let's see if we can simplify things so if we let's see if we were to if we were don't let me just rewrite everything so this is equal to X minus one over X plus five times one over X minus one minus X plus five over X minus one squared now let's think about what happens when we distribute this so when you when you distribute this times that this numerator cancels with that denominator and so we're going to get we're going to get let's see 1 over X plus five and then when you distribute it over here the X plus 5 is going to cancel with the X plus 5 and one of the X minus ones is going to cancel one of these X minus ones you're going to be left with just one of those X minus ones in the denominator and so you get F prime of X is equal to this and lucky for us we got the same answer either way but as we see the easy way was much easier than the hard way