How to take Derivatives in Calculus - Differentiation Formulas - [1-3]
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- Опубликовано: 5 окт 2024
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In this lesson, you will learn how to take a basic derivative in calculus. We learn how to take the derivative of a constant, a polynomial, and the sum and difference of constants and polynomials. We have established that the derivative is the rate of change of the function. It represents the slope of the line tangent to the original function for any given value of x. The formulas and techniques for taking basic derivatives are straightforward and easy to learn and practice.
![Derivatives of Trig Functions (Sin, Cos, Tan) in Calculus - [1-4]](http://i.ytimg.com/vi/ocuZkDGdy1Y/mqdefault.jpg)
![Derivatives of Trig Functions (Sin, Cos, Tan) in Calculus - [1-4]](/img/tr.png)
Awesome video! Thank you!
I never had the opportunity to take calculus. Thank you for the opportunity of a lifetime!
Thank you for this series! One step at a time, you've simplified a topic that otherwise is overwhelming!
Learning from your videos, a person quickly becomes a technical master.
Other vids just show shortcuts or leave out certain things the teacher thinks unnecessary.... not here.
Your teaching is excellent..❤❤❤
Thanks so much sir, we really appreciate
As always amazing!!!!
Thanks again!
The derivative of Y(x+2)⁵...❤❤❤Please it's really helpful to me Sir❤❤❤
Helpful video ! 10x sir 💕💕💕
Glad to hear that!
Nice 👍
Thanks 🙏👍 💕❤️👑
You're welcome 😊
Please more lessons trigonometric I m form four
Will do!
I will wait you thanks
Makes me cry.. math scared me
Hopefully tears of joy!
Calculate the number of ads per minute over a time period of 34:58.
Can you help me with the volume of sphere one please? Maybe it’s a silly question, but…
Pi is a constant, therefore its derivative supposed to be 0.
Then why not 4/3 x 0 x r^2 = 0
Thank you
V = 4/3 * pi * r^3, dV/dr = d/dr (4/3 * pi * r^3)
Take the constants outside the derivative: dV/dr = 4/3 * pi * d/dr (r^3) = 4/3 * pi * 3r^2 = 4*pi*r^2
It is indeed 4/3·π·r². If it were 4/3+π+r², it would be the deriv. of 4/3 + deriv. π + deriv. r². However, now it is the deriv. of 4/3·π·r² as a whole.
If it is still unclear, you can approach it two ways to possibly make it easier. Otherwise feel free to skip the following as it may make it more confusing.
You can multiply the π to the fraction 4/3 and get (4π/3), resulting in (4π/3)·r² (which holds the same value as 4/3·π·r², just rewritten). If it is still unclear, we can move the first part of the multiplication to a variable, let's say "s".
v(r) = s·r³
Before using 4π/3, if s were "5", it would be v(r) = 5r³. And we know: v'(r) = 5·3·r² = 15r²
However, s is 4π/3, so:
s = 4π/3
v(r) = s·r³
v'(r) = (or: dv/dr =) s·3·r² = (4π/3)·3·r² = [(3·4π)/3]·r² = 4π·r²
@@thepedzed yeah, sure, since that i figured out i missed the multiplying rule :)
Thanks anyway
mchew....🤣🤣🤣..all the time