This is probably the most direct extension of integer factorial one could think of. But as we see, the easy way is much easier than the hard way.So we are looking for a function that satisfies And lucky for us, we got the same answer either way. And so you get F prime of X is equal to this. And you're going to be left with just one of those X minus ones as the denominator. And one of the X minus ones is going to cancel one of these X minus ones.
Derivative of log x plus#
And then when you distribute it over here the X plus five is going to cancel the X plus five. With that denominator and so we're going to get, going to get, let's see, one over X plus five. So when you distribute this times that, this numerator cancels Now let's think about what happens when we distribute this. Minus one over X plus five times one over X minus one, minus X plus five, over X minus one squared.
So if we, let's see, if we were to, if we were to let me just rewrite everything. Or you can say negative X minus one to the negative two power times this first expression over there. Negative one over X minus one and the, over X minus one squared. And then I took the derivative of this which is this right over here. Derivative, derivative of this is one times that. So actually, so times, times, X plus five. Use inv to specify inverse and ln to specify natural log respectively Eg:1. A useful mathematical differentiation calculator to simplify the functions. And then times theĭerivative of X minus one with respect to X. An online derivative calculator that differentiates a given function with respect to a given variable by using analytical differentiation. So I can say negative one, X minus one to the negative two. Well, let's see, that's going to be, that's going to be negative X minus one to And then plus, what's the derivative of X minus one over negative one. So let me write this this is X minus one over X plus five times so let's apply the product rule, The derivative of X plus five, well that's just one times the second term times X minus one to the negative one, so that's one over X minus one. So this thing so let me rewrite it, I think you already appreciate why this is the hard way. And I like to write it that way because I always forget the whole quotient rule thing. So that's going to be times and I'm going to rewrite it as the derivative with respect to X of X plus five times X minus one to the negative one power. And then it's going to be times, and I'll do this in magenta. So let's see, this is going to be equal to, let's use some colors here, this, what I'm boxing off in blue, that's the same thing as X minus one over X plus five.
The derivative of this whole thing with respect to this expression, times the derivative of this expression with respect to X. In that case, F prime of X is going to be the derivative of this whole thing with respect to X plus five over X minus one which is going to be one over X plus five over X minus one times the derivative, times the derivative with respect to X of X plus five over X minus one. Simplify this expression using this property and just to try and power through this using the chain rule. 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. And the derivative of this, well, let's see, we're going to have a minus sign there and the derivative of the natural log of X minus one with respect to X minus one is going to be one over X minus one and the derivative of X minus one with the respect to X is just one you just multiply this by one, it doesn't really change the value. I'm just applying the chain rule here, and that's just going to be one.
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 five with respect to X plus five, so that's going to be one over X plus five times the derivative of X plus five with respect to X. We can write F of X as being equal to the natural log of X plus five minus the natural log of X minus one. It from a point of view in terms of having to So if we just apply this property right over here, and just simplify this expression, or at least simplify
So this is just going to be equal to the natural log of A minus the natural log of B. The easy way is to recognize your logarithm properties, to remember that the natural log of A over B. And I encourage you to pause this video and try to figure it out on your own. And what we want to figure out is what is F prime of X. Voiceover:Let's say that we've got the function F of X and it is equal to the natural log of X plus five over X minus one.