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### Course: Differential Calculus>Unit 3

Lesson 10: 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?
• That's one way
• Why does Wolfram Alpha give 5/2 as the answer?
• 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.
• I'm curious how you can solve this by using the limit definition. Anybody succeeded in that?
• isn't the derivative of 6arctan(x) = 6/cos^2(x)? in practice it says derivative of 6arctan(x)=6*1/(1+x^2).
• 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!)
• Is f(x)=5*log(x) true ? If yes then shouldn't be f'(x)=5/x ?
(1 vote)
• 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?
• 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)
• is the limit definition irrelevant when it comes to advanced topics?
• How do you solve it with the limit definition? Is it possible?
• You can solve most derivatives using limit definition but its not true with inverse functions.
(1 vote)
• 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.