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Differentiating logarithmic functions using log properties

By exploiting our knowledge of logarithms, we can make certain derivatives much smoother to compute. Created by Sal Khan.
Video transcript
Let's say that we've got the function f(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'(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 can approach this. I would call one way the easy way, and the other way the hard way. 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, and simplify this expression or at least simplify it from the point of view of having to take this derivative, we can re-write 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, let's see, we're gonna 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-1 and then the derivative of (x-1) with respect to x is just 1. It 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 tried to approach it on your own. Well, that's not to simplify this expression using this property and to just try to power through this using the chain rule. So let's try to do that. So in that case, f prime of x is going to be the derivative of this whole thing with respect to x plus 5 over x minus 1. So it will 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 general rule. 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. This one I'm boxing off in blue, that's the same things as x minus 1 over x plus 5. I'm just taking the reciprocal of this. And then it's gonna be times and I'll do this in magenta. That's not magenta. So that's gonna be times and I'm gonna rewrite it as a derivative with respect to x of x plus 5 times x minus 1 to the negative 1 power. 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'd appreciate it why this is the hard way. So let me write this is x minus 1 over x plus 5 times, lets apply the product rule. Derivative of x plus 5. Well that's just 1 times the second term times x minus 1 with a negative 1. So that's 1 over x minus 1 and then plus. What's the derivative of x minus one to the negative one. Let's see, that's going to be, that's going to be negative x minus one to the negative two power. So I could say negative one x minus one to the negative two, or I could just write it like this. And then times the derivative of x minus one with respect to x. Well, that's just going to be 1, and then times this, x plus 5. So actually, so times, times x plus 5. So all that I did here, product rule. Derivative, derivative of this is 1, 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, over x minus one squared, or you could say negative x minus one in 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 to, let me just rewrite everything. So this is equal to x minus 1 over x plus 5 times 1 over x minus 1. Minus x plus 5 over x minus 1 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 5. 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 1s is going to cancel with one of these x minus 1s. And you're going to be left with just one of those X minus ones as the denominator. And so you get f prime of x is equal to this, and lucky for us, we've got the same answer either way. But as we see the easy way was much easier than the hard way.