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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 28: 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
• 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.
• 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.
• 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.
• 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
• 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
• How do you derive a function that has x being raised to the power of x and something else? For example x^(tan(x))
• 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).
• I do not understand why ln(x+5) - ln(x-1) becomes 1/(x+5) - 1/(x-1) when differentiated?
• 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.
• 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).
• 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!
• 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)