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

Lesson 34: Disguised derivatives# Disguised derivatives

We tackle a complex limit expression, viewing it as 5 times the derivative of log(x) when x=2. This perspective enables us to evaluate the limit using differentiation rules, providing a simpler alternative to direct limit analysis. Together, let's navigate this challenging problem.

## Want to join the conversation?

- Instead of factoring out the fives, couldn't we have defined f(x) as f(x) = 5*log(x) and continued accordingly?(33 votes)
- That's one way(4 votes)

- Why does Wolfram Alpha give 5/2 as the answer?(4 votes)
- You have to be careful with "log". The convention on KA is to assume this means log to the base 10, AKA the common logarithm.
**This is by no means universal**. Engineers and calculator manufacturers generally agree, but mathematicians sometimes take log to mean the natural logarithm (base e).

That was the way I was taught, and I had to "unlearn" it for KA.

Wolfram Alpha is in the other camp from KA. Log means log to the base e. This is why it pops up a warning message with the answer saying:

"Assuming "log" is the natural logarithm | Use the base 10 logarithm instead".

If you click on the link in that message you get an answer that agrees with Sal.(17 votes)

- I'm curious how you can solve this by using the limit definition. Anybody succeeded in that?(4 votes)
- Is f(x)=5*log(x) true ? If yes then shouldn't be f'(x)=5/x ?(1 vote)
- The above only works if your log base is 'e'. Otherwise, you have to divide by the natural log of the the log base, which is 10 in this example. For more help on derivatives of logs, see https://www.khanacademy.org/math/calculus-home/taking-derivatives-calc/logarithmic-functions-differentiation-calc/v/logarithmic-functions-differentiation-intro(6 votes)

- In one of the practice questions we are asked to find the limit as h approaches zero of (3ln(e+h)-3ln(e))/h. I understand we can factor out the 3, and take the derivative of ln(e). The correct answer is 3/e and I understand how we get there, and looking at the graph of y=3ln(x) at x=e it makes sense.

However, I feel like I'm missing something mathematically. Since ln(e) is equal to 1, and the derivative of a constant is equal to 0, why is the derivative not equal to 3x0 or just 0?(2 votes)- You are trying to take the derivative of the function f(x)=3ln(x)

As the definition of the derivative is`f'(x)= lim_h->0 (f(x+h)-f(x))/h`

You**already have the derivative in the statement you have written above**.

All you have to do is solve the limit statement and you have the value of f'(3)(3 votes)

- isn't the derivative of 6arctan(x) = 6/cos^2(x)? in practice it says derivative of 6arctan(x)=6*1/(1+x^2).(1 vote)
- d/dx[tanx] = 1/cos^2 x

d/dx[arctanx] = 1/(1+x^2)

I encourage you to review "differentiating inverse trig functions" on khan academy.

There, you learn that the derivatives of the inverse functions are as follows -

If f(x) = sin^-1(x)

Then f’(x) = 1/√(1-x^2)

If f(x) = cos^-1(x)

Then f’(x) = -1/√(1-x^2)

If f(x) = tan^-1(x)

Then f’(x) = 1/(1+x^2)

If f(x) = cot^-1(x)

Then f’(x) = -1/(1+x^2)

If f(x) = sec^-1(x)

Then f’(x) = 1/x√(x^2 -1)

If f(x) = csc^-1(x)

Then f’(x) = -1/x√(x^2 -1)

(you actually don't learn inverses of cot, sec, and csc on Khan academy, but they are easy to derive and I decided to throw them in there just in case you were interested!)(4 votes)

- How do you solve it with the limit definition? Is it possible?(2 votes)
- You can solve most derivatives using limit definition but its not true with inverse functions.(1 vote)

- is the limit definition irrelevant when it comes to advanced topics?(2 votes)
- please can someone explain something to me, i did the limits topic on khan academy and the moved onto derivitives, i presumed that we'd be usign limits for derivitives, i am confused where limits come into anything? do we use them to take derivitives? can we not just take the derivitive and then plug in the number as that number aproaches 0

sorry but i am so confused about applying limits(1 vote)- Derivative is defined as a limit:

𝑓 '(𝑥) = lim ℎ→0 (𝑓(𝑥 + ℎ) − 𝑓(𝑥))∕ℎ

This definition is then used to show various derivative rules, like the power rule or the product rule, but once we have established these rules we never really use the definition again, with the result that many calculus students simply forget that it ever existed.(3 votes)

- at2:01, we can also set f(x)=5log (x) and get the same result finally, right?(1 vote)
- Yep! You can do that!(3 votes)

## Video transcript

- [Voiceover] Let's see
if we can find the limit as h approaches zero of
five log of two plus h minus five log of two, all of that over h. And I'll give you a little bit of a hint because I know you're
about to pause the video and try to work through it. Think of your derivative properties, especially the derivatives
of logarithmic functions, especially logarithmic
functions in this case with a base 10. If someone just writes
log without the base, you can just assume that that
is a 10 right over there. So pause the video, and see
if you can work through it. Alright, so the key here
is to remember that if I have, if I have f of x, let me do it over here, I'll do it over here. F of x and I want to find f prime of, let's say f prime of some number, let's say a, this is going to be equal to the limit as x, or sorry, as h approaches zero is one
way of thinking about it, as h approaches zero of f of a plus h, minus f of a, all of that over h. So this looks pretty close
to that limit definition, except we have these fives here. But lucky for us we can
factor out those fives, and we could factor them out, we could factor them out out front here, but if you just have a scale
or times the expression, we know from our limit properties we can actually take those
out of the limit themselves. So let's do that, let's take both of these fives, and factor them out, and so this whole thing
is going to simplify to five times the limit as h approaches zero of log of two plus h, minus, minus log of two, all of that over h. Now, you might recognize
what we have in yellow here, cause think about it, what this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually let's say f prime of two is, well this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see there, this by definition this right over here is, is f prime of two. If f of x is log of x, this is f prime of two, f prime of two. So can we figure that out? If f of x is log of x, what is f prime of x? F prime of x, we don't need
to use the limit definition. In fact the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions. So f prime of x is going to be equal to one over the natural log of our base, our base here we already
talked about that, that is 10. So one over natural log of 10, times, times, times x. If this was a natural log, well, then, this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base you put the natural log of that base right over here in the denominator. So what is f prime of two? F prime of two is one over the natural log of 10, times two. So this whole thing has simplified, this whole thing is equal
to five times this business. So I could actually just write it as it's equal to five over, five over the natural log of 10, natural log of 10, times two, I could have written it
as two natural log of 10s. The key here for this type of exercise, you might be a little let me see if I can evaluate this limit? You're like whoa, this looks
a lot like the derivative of a logarithmic function, especially the derivative
when x is equal to two, if we could just factor these fives out. So you factor out the fives, and say hey, this is,
this is the derivative of log of x when x is equal to two, and so we know how to take
the derivative of log of x. If you don't know we have
videos where we prove this. Where you take the
derivatives of logarithms with bases other than e, and you just use that to actually and you find the derivative, and you evaluate it at two, and then you're done!