Derivative of Logarithmic Functions
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- Опубликовано: 6 окт 2024
- This calculus video tutorial provides a basic introduction into derivatives of logarithmic functions. It explains how to find the derivative of natural logarithmic functions as well as the derivative of log functions. You need to be familiar with the chain rule for derivatives. This video contains plenty of examples and practice problems.
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...
_________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Derivatives - Formula Sheet:
bit.ly/4dThzf1
Derivatives - Formula Sheet: bit.ly/4dThzf1
Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections
Next Video: ruclips.net/video/9z1Dz60mWcQ/видео.html
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Hi Everybody. Time mark 8:47 the answer 2 / X LN(4) can be simplified. Remember that 4 is the same as 2^2 and so LN(4) is the same as LN(2^2) which is the same as 2LN(2) and so 2 / X LN(4) is equal to 1 / X LN(2). Graph them and they overlap.
waw
wont it be 2x/xln4 ? I thought it is u'/ulna
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at 4:57 why don't you subtract ^1/7 - 1 ?
I'm a litfle confused on your 6th example. Why didn't you subtract 1 from the exponent 1/7 after moving it on the front???
I’m pretty sure that has to be a mistake, I looked up the derivative and it’s not that answer.
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5:03 is the wrong answer :( its 1/7x^6/7(7root of x )
Both answers are correct, because x^6/7 * x^1/7 (or 7root of x as you write it) = x^7/7 = x. So 1/7x is a correct answer, just more simplified than yours.
@@minortooth5744 um if you simplify 1/7x^6/7(7root of x ) you don't get 1/7x... you get 1/ (7 * (7root x^5))............ 7rootx can't take out the whole 7rootx^6 but can take out one factor since there's only one on the numerator.
the guy just missed the step of doing 1/7 - 1 when he initially brought the 1/7 down using the power rule
if im mistaken pls let me know
@@moo47692 I understood his notation like this:
www.symbolab.com/solver/step-by-step/%5Cfrac%7B1%7D%7B7%20x%5E%7B%5Cfrac%7B6%7D%7B7%7D%7D%5Ccdot%5Csqrt%5B7%5D%7Bx%7D%7D
I thought that by typing (7root of x) in the columns after x^6/7 he meant to multiply both expressions (when there is no sign between two numbers (like + - /) it means they should be multiplied ....... 3x(-7x) = -21x^2 etc.)
Now you can rewrite (7root of x) as x^1/7 and if you multiply two numbers like this their exponents add together. So 1/7 + 6/7 = 7/7 = 1 and x^1 = x
You would get (7root x^5) if you had (x^6/7)/(x^1/7) which I just don't see there.
Although I may have gotten that wrong, because I'm not really used to reading math equations written in 1 single line on yt, it's kinda messy :D
But it made sense for me to assume that there is just the number 1 in the numerator and the rest is in the denominator. The only other option I saw was (1/7x^6/7)*(x^1/7) = x/7
I hope I answered well, I'm slowly getting lost in what I wrote at this point :D
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At 5:03, why didn't you subtract the exponent (1/7) by 1
Did you find out why? Cause I'm lost here
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we write log base 3 x = log base 3 e* log base e x
now if we find the derivative of the expanded function. it becomes
ddx ( log base 3 e* log base e x) , log base 3 e is constant
so the statement becomes log base 3 e * ddx log base e x.
now we can easily differentiate log base e x as 1/x.
thus the entire thing becomes (log base 3 e / x).
log base 3 e = 1/ log base e 3.
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8:46 if I did not move the ^2 to the front of the ln, I would get a different answer since x^2 will be in the denominator. Am I missing something?
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If you did not move the ^2 to the front, then you would have had to use the chain rule and take the derivative of x^2, which is 2x, and multiply it by 1/(x^2*ln(4)). Notice that an x would cancel from the top and bottom and you would still get the same answer. Hope this helps!
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Thanks for the great video, but In the last example I think the correct answer would be 1/[ln(x)*x^2] , I did it using the chain rule. I think it's because u is a function of X, so you have to use chain rule even with your substitution.
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