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### Course: Class 12 math (India)>Unit 5

Lesson 15: Logarithmic functions differentiation

# 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.

## Want to join the conversation?

• Why would anyone want to do it the "hard" way, the easier way seems more understandable
(6 votes)
• Some of the hard ways of solving problems allow us to understand how they were discovered. If we look at some harder versions of solutions like the one shown in the video we get a deeper understanding of what might will come in the future.
(9 votes)
• Silly question, but in terms of simplifying the expression, is 1/(x+5) - 1/(x-1) "more simplified" than -6/(x+5)(x-1) or -6/(x^2 +4x -5)? What makes an expression simpler than another ? How easy it is to plug x (1st one)? Having it all under one group (2nd expression)? or expanding everything (3rd one) ? Again, silly question but i'm curious.
(6 votes)
• Technically, the first version of the expression is more simplified, because you cannot cancel anything from it. In your second examples, with the addition of the rational expressions, you've created something that could be simplified further through factoring and cancelling. This often depends on the teacher, but a simplified expression is generally something from which nothing can be factored and cancelled.
(3 votes)
• i used the quotient rule to derive x+5/x-1 then multiplied it by 1/x+5/x-1 and got -6x+6/x^3+3x^2-9x+5 as my final answer........ im confused
(3 votes)
• why?
u r correct.
Indeed, u should just reduce ur equation.
u have got: (-6x+6)/(x^3+3x^2-9x+5).
We can convert it into: -6(x-1)/(x^3+3x^2-9x+5).
(x^3+3x^2-9x+5)=> This is a cubic polynomial. I hate factorizing cubics but, this one is quite easier.
(x^3+3x^2-9x+5)=x^2(x-1)+4x(x-1)-5(x-1)
=>-6(x-1)/(x^3+3x^2-9x+5)= -6(x-1)/(x-1)(x^2+4x-5)
= -6/(x^2+4x-5)
Same as Sal!

If u want to factorize some tough cubic equations, try this trick out (optional):
http://www.sosmath.com/algebra/factor/fac11/fac11.html
(4 votes)
• How do you derive a function that has x being raised to the power of x and something else? For example x^(tan(x))
(3 votes)
• You can write x as e^(ln(x)): x^(tan(x))=(e^(ln(x))^(tan(x))=e^(ln(x)*tan(x)). Now you have the usual: the derivative of e^f(x) is f'(x)*e^f(x).
(4 votes)
• I do not understand why ln(x+5) - ln(x-1) becomes 1/(x+5) - 1/(x-1) when differentiated?
(2 votes)
• The derivative of ln(u) is u'/u. In this case, u for ln(x + 5) is x + 5. The derivative of x + 5 is 1. Therefore you could plug in u' and u to get 1 / (x + 5). For the derivative of ln(x - 1), u would be equal to x - 1. The derivative of x - 1 is 1, so the derivative of ln(x - 1) is 1 / (x - 1). Combining these you get 1 / (x + 5) - 1 / (x - 1). Hopefully this makes more sense and feel free to comment back with more questions.
(5 votes)
• At why is -1 in the numerator?
(1 vote)
• It follows from the power rule. Recall that if we define a function ƒ: R \ {0} -> R by ƒ(x) = x^(-1), then the derivative of ƒ at x ≠ 0 is given by (-1) · x^(-2). Now apply this to the function given by ƒ(x - 1).
(5 votes)
• Does Sal have a proof in any video that proves ln(b^a) = a * ln(b) ?
(1 vote)
• One of the exercises is y=(x+3)^3(x-4)^2 and the answer is y(3/(x+3)+2/(x-4)). When I expanded the problem I got y= x^5+x^4-29x^3-45x^2+216x+432. The derivative is 5x^4+4x^3-87x^2-90x+216. Where is my error? Thanks for your help!
(2 votes)
• The first "answer" you are giving is what you get if you take the log of both sides and then differentiate only the right hand side. Differentiating the left hand side (ln(y)) would give you 1/y * y'. Multiply both sides by y to solve for y'. Since y = (x+3)^3 * (x -4)^2, you get y' = 3(x+3)^2 * (x-4)^2 + 2(x - 4) * (x + 3)^3, which, when expanded and simplified, should give you the same result you got by expanding first and then differentiating (though I admit I didn't check).
(1 vote)
• what is difference between ln(x/y)and lnx/lny?
(1 vote)
• Using logarithm rules, Ln(x/y) = Ln(x) - Ln(y) which is different from Ln(x)/Ln(y)
(2 votes)
• Why and/or when would I ever need to approach a problem like this the "hard way"?
(1 vote)
• Because you haven't spotted the easy way. The point is, that even though we can sometimes take shortcuts in maths, we arrive at the same (or equivalent) conclusions no matter which way we proceed, as long as we don't make any mistakes along the way.

Also, the way I did it I actually came up with a different form of the result, which may be more useful depending on what you want to do - I used the quotient rule, and ended up with -6/((x+5)(x-1)) which expands to -6/(x^2+4x-5). Even though that's equivalent to Sal's solution, it's not trivial to get from one form or the other. Note that Wolfram Alpha also gives "my" form when asked to differentiate ln((x+5)/(x-1)).
(2 votes)

## Video transcript

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. 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 can 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, and just simplify this expression, or at least simplify it from a point of view in terms of having to take this derivative. We can rewrite F of 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. 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. I'm just applying the chain rule here, and that's just going to be one. 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 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. 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 just to try and power through this using the chain rule. So let's try and do that. 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. This is just the chain rule. The 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'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. I'm just taking the reciprocal of this. And then it's going to be times, and I'll do this in magenta. No that's not magenta. 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 I like to write it that way because I always forget the whole quotient rule thing. But I remember the product rule. So this thing so let me rewrite it, I think you already appreciate why this is the hard way. 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. And then plus, what's the derivative of X minus one over negative one. Well, let's see, that's going to be, that's going to be negative X minus one to the negative two power. So I can say negative one, X minus one to the negative two. I can just write it like this. And then times the derivative of X minus one with respect to X. Well that's going to be one. And then times this. X plus five. So actually, so times, times, X plus five. So all I did here, product rule. Derivative, derivative of this is one times that. And that gave us that over there. And then I took the derivative of this which is this right over here. Negative one over X minus one and the, over X minus one squared. Or you can say negative X minus one to the negative two power times this first expression over there. So there's that derivative. 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 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 distribute this times that, this numerator cancels with that denominator and so we're going to get, going to get, let's see, one over X plus five. And then when you distribute it over here the X plus five is going to cancel the X plus five. And one of the X minus ones is going to cancel one of these X minus ones. 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 got the same answer either way. But as we see, the easy way is much easier than the hard way.