Introduction to Logarithmic Differentiation
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- Опубліковано 26 лют 2018
- This calculus video tutorial provides a basic introduction into logarithmic differentiation. It explains how to find the derivative of functions such as x^x, x^sinx, (lnx)^x, and x^(1/x). You need to take the natural log of both sides of the equation and perform implicit differentiation to find dy/dx. You need to be familar with product rule for derivatives and properties of logarithms.
Derivatives - Fast Review:
• Calculus 1 - Derivatives
Derivatives - The Product Rule - f*g:
• Product Rule For Deriv...
Derivatives - The Quotient Rule:
• Quotient Rule For Deri...
Derivatives - The Chain Rule:
• Chain Rule For Finding...
Derivatives - Composite Functions:
• Derivatives of Composi...
__________________________________
Implicit Differentiation:
• Implicit Differentiation
Derivatives of Inverse Trig Functions:
• Derivatives of Inverse...
Derivatives of Exponential Functions:
• Derivatives of Exponen...
Derivatives of Logarithmic Functions:
• Derivative of Logarith...
Logarithmic Differentiation:
• Introduction to Logari...
___________________________________
Derivatives - Using Logarithms:
• Finding Derivatives Us...
Derivatives of Inverse Functions:
• Derivatives of Inverse...
Derivatives - Differentiation Rules:
• Basic Differentiation ...
Derivatives - Function Notations:
• dy/dx, d/dx, and dy/dt...
Derivatives - The Reciprocal Rule:
• The Reciprocal Rule an...
_________________________________
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Shortcut for derivative of a function of type: f(x) raised to g(x)
Let f(x) be f
Let g(x) be g
Let derivative of f(x) be f'
Let derivative of g(x) be g'
d/dx(f^g)=(f^g)(f'(g/f)+g' log(f))
For those who couldn't understand what's written above, It reads as:
derivative of f raised to power g is equal to f raised to power g times (derivative of f times g by f + derivative of g into log(f))
ok Harvard material.
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It has really helped me understand logarithmic differentiation
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Once you get out of school, logarithmic differentiation can be used in solving problems that traditionally require the product rule or the quotient rule. Less formulas to memorize!
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In the 3rd example, shouldn't the variable exponent go inside the parenthesis, and then bring it down in the parenthesis of the outside most ln
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Excuse me sir..please how come you got
1- 1nx after using the product rule... I'm a little bit confused
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Anyone can please answer me, at 1:06 , when taking derivatives of ln y, why 1/y has to be further multiplied by dy/dx? whats the use of the dy/dx here? Any difference if we just keep 1/y? Thankyou!
If you needd an explanation, search implicit differentiation!
you are applying d/dx to y, if it was d/dx to x then it would just be 1/x
if you apply d/dx to anything other than x you have to tag a d(variable)/dx to the end
dy/dx is saying the derivative of y with respect to x
since y=f(x), ln(y) is the same as ln(f(x)), so it's a compound function and we need to use the chain rule. the derivative of ln(y) is 1/y and the derivative of y is dy/dx
5:54
can I take take the derivative of [ln x^x] like 1/x^x × derivative of x^x?
Thank you sir. :))
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4:00 why the derivative of ln y = 1/y*dy/dx? why dy/dx?
I am not sure, but when it comes to the derivative of anything related to y, it should be multiplied by dy/dx.
Because we have defined y to be a function of x. So, because y is a function, we have to use the chain rule when we differentiate ln(y).
Hey,in the first example where he showed d/dx(x^x), isn't Ln x a transcendental function? Then why can we take the derivative of both sides
Great video, but why do you have to take the natural log at 6:14? Why can't you just bring down the exponent immediately?? Any response would be greatly appreciated
@Giorno Giovanna I see, but my confusion is coming from why you can't bring down the exponent as is ln(x)^x --> xln(x) instead of the procedure shown in the video.
@@SquatSimp Well, to answer that question, I have to ask you one first: why do you think you _can_ do that?
If it said ln(x^x), you'd be correct, that can be written as xln(x). But it *doesn't* say that, it says ln(x)^x, which has nothing at all to do with xln(x).
At the 10:10 why we do not use the quotient rule?
You can use the quotient. It will just end up the same answer.
I have a question. At 8:17 you use the natural log derivative formula and use u'/u to get 1/x ln x . But I used the chain rule formula f(g(x)) = f ' (g(x)) ⋅ g'(x) and got the same answer. Is this valid? Thank you :)
Yes that's valid..even, if u use implicit func basic formula that's give the same ans. too
To be honest you don't need to add the u' in every derivative techniques as long as you understand how composition of function works, where in theory every techniques then actually uses the chain rule.
for the second question, can't we just take the x to the front? cuz that's an ln property? 6:00 and that gives us x(1/x) which is one?
Why do we use ln x instead of ln |x|, so there are less restrictions on the f'(x) ?
with absolute value there's gonna be more mess
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why can't we apply the chain rule for the problem: d/dx [ lnx ] ^ x
If you want to see how to make things clear, you can rewrite (ln(x))^x using base e as (e^ln(ln(x)))^x=e^(xln(ln(x))). Then, differentiate using chain and product rule. You would end up as e^(xln(ln(x)))(x(1/(xln(x)))+ln(ln(x))))=(ln(x))^x(1/ln(x)+ln(ln(x))), which is the same answer as before.
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Question: y= (sin x)^logcosx^2
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At 9:38 why do you choose to take the natural log instead of taking a normal log?
The derivative of ln is simpler than the derivative of other logs but you could use any log base you want and get an equivalent answer
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variables = letter numbers
Couldn't you simplify sin(x)/x to 1?
No, this is math, not engineering
6:05 why did you insert another natural logarithm there?
savior
Can someone explain d/dx(ln(lnx)) is not 1/Inx? Like lny become 1/y. thanks
For the last one, couldn’t you also answer it as
((1-lnx)/x^2)(x^1/x)?
No
@@lalotothemax Why not?
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Wonderful explanation
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Can anyone explain the lnx in 8:33
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For the first problem, can't you simplify sin x/x and write it as 1?
Bro you can do that when x is tending to zero.. in limits basically.
Why at 8:06 do we multiply the top and bottom by x?
To simplify it i think
@@user-hv6ef9ie1gyess it would be easier after simplifying else you would have to apply quotient rule..