Derivatives of Exponential Functions & Logarithmic Differentiation Calculus lnx, e^2x, x^x, x^sinx
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- Опубликовано: 11 сен 2016
- This calculus video tutorial shows you how to find the derivative of exponential and logarithmic functions. it also shows you how to perform logarithmic differentiation. It contains plenty of examples and practice problems that involve differentiation techniques such as the power rule, product rule, quotient rule, and the chain rule.
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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@@mohanranaeducationclasses3166 help me please at 27:18 i don't understand why he cancel the x on the top
@@sanopann1896 It was to simplify the answer. Since there was a fraction in the numerator, he multiplied top and bottom by x to get rid of the fraction "1/x"
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big sis help me please at 27:18 i don't understand why he cancel the x on the top
@@sanopann1896 hey i hope you got help w this, but basically he wants to simplify it, & there's this thing with fractions where you can multiply the top and bottom by anything (as long as you multiply the same thing in the top and bottom, it doesn't change it at all. it's like multiplying by 1/1). so you can choose something to conveniently make the fraction easier to work with, like this. when you do [(1/x)*(x/1)] you end up with 1*x/x*1... simplifies to x/x, or 1 :) so then the whole problem is easier to manage
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Trigonometric Derivatives
*Hope this helps!*
d(ln (u)) = (1/u) (du/dx)
d(log_a^u) = (1/ln a u) (du/dx)
d(e^u) = e^u du/dx
d(a^u) = ln a a^u (du/dx
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big bro need help at 5:12 why it was ex^e-1 instead of ex
@sano pann because it's variable raised to a constant, the derivative of x^2 is 2x. The derivatives of 2^x is (2^x)(1)ln2
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