MAT183 VIDEO ASSIGNMENT: TECHNIQUES OF DIFFERENTIATION

Añadido: 2026-06-02

[00:00:02][music] [bell] [music] >> What is differentiation?

[00:00:15]>> Differentiation is a concept in calculus that is used to find the rate of change of a function with respect to its variable. It shows how quickly a quantity increase or decrease at a particular point and determine the slope of the tangent line to a curve. The result obtained from differentiation is called the derivative, which represent the instantaneous rate of change of a function. Now, we are going to look at different technique of differentiation.

[00:00:36]The first one is constant rule, which is the derivative of the constant is zero.

[00:00:39]>> [music] >> The next is power rule, which is used to differentiate the power of X. Product rule is used when two function are [music] multiplied together.

[00:00:47]Quotient rule is when the function is divided by another function. [music] For sum and difference, the sum is plus while the difference is minus. Chain rule is applied when a function is inside another function. Trigonometric is to differentiate trigonometric function. Logarithm is used to differentiate logarithmic function.

[00:01:03]Exponential is to differentiate exponential function. Implicit differentiation is to differentiate function of Y with respect to X.

[00:01:10]>> My question two. The question is, find dy/dx for the following function. The first equation is y = ln x sec 2x. Now, I will show you how to solve this equation. Uh first, I will divide this equation by two. The first one will become U and the second one will be V.

[00:01:23]So, rewrite the U again.

[00:01:26]U equal to ln x cubed.

[00:01:29]So, we all know ln to differentiate it will be 1 over x. So, U equal to 1 over x cubed. And then, we need to differentiate this one. It will be like this.

[00:01:40]3 x squared.

[00:01:41]And then, we need to find this. It will be like this.

[00:01:45]3 over x.

[00:01:49]That's it. And then, for the V, V equal to sec 2x. And then, we differentiate V.

[00:01:56]So, we all know sec if we want to differentiate, it will be secant x tangent x. So, secant write this again, secant 2x tangent 2x. And then, we need to differentiate this. It will be 2.

[00:02:09]Okay, after we differentiate both of the equation, we should apply product rule, which is dy over dx equal to uv' + vu' which is ln x cubed times 2 secant 2x tangent 2x plus v secant 2x times u' 3 over x And then we swap these. They'll be Put this two in the front. So, it'll be 2 ln x cubed secant 2x tangent 2x plus This one, bring to front.

[00:02:47]3 over x secant 2x That's it. Okay, the next question is y = 3 - x squared over exponent tangent x. We need to rewrite the equation.

[00:02:57]It'll be like this. y = 3 - x squared and then in the bracket e minus tangent x.

[00:03:04]And then, we can divide the equation same like the equation that we have done before, which is this one will be u and this one will be v. And then, we write it again. u = 3 - x squared and then u' will be only differentiate this one. It'll be negative 2x. And then, v rewrite it again, exponent negative tangent x.

[00:03:24]v' We only If you want to differentiate exponent, we need to rewrite the exponent first and then differentiate the x. So, it'll be exponent negative tangent x and then tangent will be differentiate it. It'll be negative secant. Okay, after we get all of the equation, we need to apply product rule, which is dy over dx equal to u'v + v'u. And then, we need to substitute all of these in this equation. It'll be >> [snorts] >> dy over dx equal to -2x times exponent negative tangent x plus b prime exponent minus tangent x negative secant 2x and then times a times 2 3 minus x squared And then there is two same equation. We need to factorize it out. In this equation the exponent was the same. So, it will be like this.

[00:04:12]e exponent minus tangent x and then derive all of the equation.

[00:04:15]minus 2x plus negative secant squared x and then 3 minus x squared. And then we can solve it. It will be e minus tangent x negative 2x plus minus will be minus. So, minus but we can bring this to the front. It will be 3 minus x squared and then secant 2x.

[00:04:32]Now, we can see the exponent is negative. To make it positive it should be like this.

[00:04:36]x minus 3 minus x squared and then secant 2x over exponent tangent x.

[00:04:42]That's it.

[00:04:44]>> Hi. Today I would like to explain question three. It state given 3y squared plus ln x over y squared plus sine exponential with the power of x minus 2x equal to 10. Find dy over dx using implicit differentiation.

[00:05:02]First, [music] we need to rewrite the equation like this.

[00:05:08]Then [music] we need to simplify this equation through changing ln part by obeying its properties. [music] Then we can differentiate this equation by using implicit differentiation by applying the chain rule with respect to x. Let the answer of dy over dx remain [music] as its position and bring another to the other side. From this equation [music] we can factorize dy over dx. Then let dy over dx alone.

[00:05:43]>> [music] >> We can simplify this equation by multiplying it with X and Y distinctively.

[00:05:52]And [music] we can get final answer like this. dy over dx equal to negative Y minus xy >> [music] >> exponential with the power of x minus 2 times with [music] cos exponential with the power of x minus 2x over 6xy [music] squared minus 2x.

[00:06:16]>> Okay, I will discuss about question four. Question four say use linear approximation to estimate the value of 7.98 squared minus 1 over cubic root 7.98.

[00:06:26]Give your answer correct to four decimal place.

[00:06:28]Before Before that, I will write formula for linear approximation.

[00:06:33]fx approximately to fx plus fx fx prime x minus x.

[00:06:45]We know 7.98 is value for x.

[00:06:49]So, x equal to 7.98.

[00:06:53]After this, we can find x.

[00:06:56]x not must always be integer and always close to x value. So, I will choose 8 as my x not. Okay, after this, we can let 7.98 become x.

[00:07:14]So, this equation replace x into this equation.

[00:07:20]fx equal to x squared minus 1 over cubic root x.

[00:07:28]After we find fx, we can find fx prime.

[00:07:32]By simplifying this equation.

[00:07:35]fx prime equal to 2x minus minus 1 over 3 x to the power of -4/3.

[00:07:44]But, we don't like the power is negative.

[00:07:47]So, we must simplify.

[00:07:49]2x minus minus become positive.

[00:07:53]1/3 x to the power of 4/3.

[00:07:58]Okay, after we find f prime, we can substitute it into this formula.

[00:08:03]So, f x approximately to f x not >> [snorts] >> plus f x prime times x minus x not.

[00:08:15]So, just substitute.

[00:08:17]7.98 plus f prime 8 times 7.98 minus 8.

[00:08:32]This After this, we need to substitute into this f x.

[00:08:37]Plus substitute 2x f prime 2x plus 1/3 8 to the power of 4/3.

[00:08:52]And times [clears throat] -0.02.

[00:08:56]Why is it 0.02? Because 3 minus 7.98 minus 8, we get -0.102.

[00:09:01]After this, we can simplify.

[00:09:04]We get 64 minus 1/2 plus 2 times 8, we get 16.

[00:09:11]Plus 1/3 times 8, we get 48.

[00:09:17]Times 3 -0.02.

[00:09:21]After this, we can simplify again.

[00:09:24]64 minus 1/2, we get 63.5.

[00:09:28]Plus we get this one 16 plus 1/48 times 0.02.

[00:09:40]Then, we can calculate all this.

[00:09:43]We get 63.1796.

[00:09:47]This is the final answer.

[00:09:52]>> [music]