Question 1

Could I instead of replacing y=ln(x) on 1/(e^y), replace e^y=x on 1/(e^y)? Is there any problem when doing that? It seemed more simple than the way Sal did it.

Accepted Answer

There are usually about 42 ways of approaching a problem.  The way you described is simpler, but sometimes Sal seems to enjoy the scenic route, so to speak.

Question 2

Is there a video where he covers why the derivative of e^y is e^y * dy/dx? I thought it would just be e^y?

Accepted Answer

Well, worth remembering our y is a function of x. If you have y as a constant value (i know `y` wouldn't be the best letter to be used for a constant, but just in case), you can say that d(e^y)/dx = e^y.

Question 3

Say you had a constant in front of the ln x. Would you just multiply the derivative of ln x by the constant?

Accepted Answer

Yes, that is correct.  For example,  d/dx  of  5 ln x = 5/x
 
Note, however,  that the d/dx of  ln (ax)  (where a is constant, not a variable) is a bit tricky.  
d/dx ln (ax) = 1/x.
Here's proof:
y = ln (ax)
let u = ax
thus du/dx = a
y = ln u
dy/dx = d (ln u)/du  ∙  du/dx
dy/dx = [1/u ]  ∙  du/dx
back-substituting
dy/dx =  [1/(ax)]  ∙  a
 dy/dx = 1/x

Question 4

what is the first derivative of y=e^-x ln^x

Accepted Answer

It's simple. You just need to know the rules. So first, take the first derivate of the entire thing. You'll get y' = (e^-x)' * (ln x) + (e^-x) * (ln x'). If you simplify this using derivative rules, you'll get y' = (e^-x * -1) * (ln x) + (e^-x) * (1/x).
Hope this helps! If you have any questions or need help, please ask! :)

Question 5

Would the same process be applied to a variable that is raised to the natural log, such as y= x^lnx ?

Accepted Answer

That is a bit more complicated. Here is the derivative of that:
```y = x^ln x
let u = ln x
thus, x = e^u
thus, du = dx/x
thus,
y= x^ln x = (e^u)^u = e^(u²)
Let w = u²
Thus, dw = 2u du
e^(u²)= e^w
Taking the derivative:
dy = e^w dw
dy = e^(u²) (2u)(du)
dy = (x^ln x)(2 ln x)(dx/x)
dy =  (2 ln x) x^[ln (x) - 1)]dx
dy/dx =  (2 ln x) x^[ln (x) - 1)]```

Question 6

i understand all that you said. so what would be the derivative of log( ln x)?

Accepted Answer

Firstly log(ln x) has to be converted to the natural logarithm by the *change of base formula* as all formulas in calculus only work with logs with the base e and not 10. 
Hence log( ln x ) = ln( ln x ) / ln (10) and then differentiating this gives [1/ln(10)] * [d(ln(ln x)) / dx]. 
This can be differentiated further by the *Chain Rule*, that gives [1/ln(10)]*{ [1/ln(x)*1/x ].
Hence the result is ( 1 / [ln(10)*ln(x)*x] )

Question 7

2:29 How does e^lnx simplify to x?

Accepted Answer

ln(x) is defined as the inverse function of e^x. If you compose inverse functions, you get back the original input, x.

Question 8

Can you use the chain rule to get the derivative of the natural log of the natural log of x? i.e ln(ln(x)). I saw this in a practice problem in composite exponential differentiation and tried to apply the chain rule, but the solution was wrong. So is the chain rule not applicable in this situation?

Accepted Answer

The chain rule applies in any situation where you have composed functions, and ln(ln(x)) is no exception. The chain rule gives you
d/dx ln(ln(x))
=1/(ln(x))·d/dx ln(x)
=1/(ln(x))·(1/x)
=1/(x·ln(x))

Question 9

At about 1:20, why is d/dx of e^y = e^y * (dy/dx) ? I thought that by the power rule, d/dx should be y * e^(y-1) ... is it because d/dx is being taken of something other than x?

Accepted Answer

There are two reasons why what you said isn't true:
1) the derivative of e^x is e^x not xe^x-1
2) when your taking the derivative with respect to x of something that has a y you must apply the chain rule and take the derivative of the outer function (in this case e to the something.) with respect to that something. so you take d/dy of e^y first which gets you e^y, then you multiply by d/dx of the inner function which is y. we don't know what the derivative of y with respect to x is so we just write it as dy/dx, an unknown that you can solve for to figure out. so the answer is indeed e^y*dy/dx.

AP®︎ Calculus AB (2017 edition)

Course: AP®︎ Calculus AB (2017 edition) > Unit 4

Derivative of ln(x) from derivative of 𝑒ˣ and implicit differentiation

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