movieboxappsxyz

★ 020206h 11m

CALCULUS

Matematika

Differentiating a function, A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. Example Consider the straight line y = 3x + 2 We can calculate the gradient of this line as follows. We take two points and calculate the change in y divided by the change in x. When x changes from −1 to 0, y changes from −1 to 2, and so No matter which pair of points we choose the value of the gradient is always 3. y is a function of x, so y =f(x) f(x1) = y1 and f(x2) =y2 x2-x1=dx, x2= x1+dx, By substitituting these terms in slope formula, we have; m =[ f(x1 + dx) - f(x1)]/dx, as the limit of dx approaches 0 Seven steps to avoit the spread of Corona Virus (Covid19): If you find this video interesting, kindly subscribe to my channel for more exciting Maths tutorials. Subscribe link: #Differentiation #First #Principles WhatsApp group: Facebook: Instagram: Linkedin: Blog:

Title overview

Core release details, quality markers, and stream status stay together here instead of floating below the fold.

Matematika

Tahun

2020

Durasi

6h 11m

IMDb

0

Playback

Start watching now

Release2020

Runtime6h 11m

IMDb★ 0

Tonton Sekarang

Mungkin Anda Juga Suka

Primary 3 Mathematics

Matematika

Montessori Math Fun for 1st Graders! 🎲🌟

Matematika

رياضيات منهج مصري ثاني ابتدائي 2015 - 2016 الترمين

Matematika

رياضيات للصف الرابع الابتدائي الترم الأول المنهج الجديد 2022

Matematika

Pry2 Math

Matematika

Grade 11 Mathematics

Matematika

JSS 3 Junior Mathematics

Matematika

Mathematics (Junior Secondary School 1)

Matematika

SSS12 Math

Matematika

SSS 1 Mathematics

Matematika

Mathematics Lessons for JAMB, WAEC, NECO, NABTEB, SS1, SS2 and SS3 students

Matematika

SS3 Mathematics

Matematika

Celebrating STEM: counting and multiplication

Matematika

SSS 2 Mathematics

Matematika

Pry1 Math

Matematika

SSS 3 Mathematics

Matematika

SSS10 Math

Matematika

SSS11 Math

Matematika

Pry 1 Primary Mathematics

Matematika

JSS7 Math

Matematika

PrePrimary Maths

Matematika

Year 11 Math

Matematika

Year2 Maths

Matematika

Video 2 Bangun Ruang

Matematika

Pilihan Unggulan

Ingin hiburan tambahan? Jelajahi hub game rekomendasi kami dan destinasi short drama pilihan.

Game Instan

Gamixo

Mainkan game browser kasual seperti puzzle, kartu, olahraga, dan tantangan cepat dalam hitungan detik.

Short Drama

VSKIT Drama

Tonton mini-series trending, kisah romansa, dan short drama terbaru dengan update harian cepat.

Komentar

10 Komentar

IMVU_jxt_•Oct 22, 2025

Integration by partial fractions is an integration technique which uses partial fraction decomposition to simplify the integrand. The integrand is written as partial fractions and then evaluated using standard methods. The integrals of many rational functions lead to a natural log function with absolute value expressions. This video explains what to do when you have non repeated linear factors factors.

EwurafuaOct 22, 2025

Integration by partial fractions is an integration technique which uses partial fraction decomposition to simplify the integrand. The integrand is written as partial fractions and then evaluated using standard methods. The integrals of many rational functions lead to a natural log function with absolute value expressions. This video explains what to do when you have non repeated linear factors factors.

Dianellisse RimaOct 22, 2025

To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

khelly Oct 22, 2025

To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

provoicelameckOct 22, 2025

Consider a differential equation of type y′′+py′+qy=0, where p,q are some constant coefficients. For each of the equation we can write the so-called characteristic (auxiliary) equation: k2+pk+q=0. The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options: Discriminant of the characteristic quadratic equation D is greater than 0. Then the roots of the characteristic equations r1 and r2 are real and distinct. In this case the general solution is given by the following function y(x)=C₁eʳ¹ˣ+C₂eʳ²ˣ, where C1 and C2 are arbitrary real numbers. Discriminant of the characteristic quadratic equation D=0. Then the roots are real and equal. It is said in this case that there exists one repeated root r of order 2. The general solution of the differential equation has the form: y(x)=(C₁x+C₂)eʳˣ. Discriminant of the characteristic quadratic equation D is less than 0. Such an equation has complex roots

🇲🇼Tik Tok Malawi🇮🇳🇲🇼Oct 22, 2025

Consider a differential equation of type y′′+py′+qy=0, where p,q are some constant coefficients. For each of the equation we can write the so-called characteristic (auxiliary) equation: k2+pk+q=0. The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options: Discriminant of the characteristic quadratic equation D is greater than 0. Then the roots of the characteristic equations r1 and r2 are real and distinct. In this case the general solution is given by the following function y(x)=C₁eʳ¹ˣ+C₂eʳ²ˣ, where C1 and C2 are arbitrary real numbers. Discriminant of the characteristic quadratic equation D=0. Then the roots are real and equal. It is said in this case that there exists one repeated root r of order 2. The general solution of the differential equation has the form: y(x)=(C₁x+C₂)eʳˣ. Discriminant of the characteristic quadratic equation D is less than 0. Such an equation has complex roots

_𝘯𝘢𝘫𝘶𝘭𝘪𝘢❤️‍🔥Oct 22, 2025

To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

THECUTEABIOLAOct 22, 2025

To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

Sonica RokayaOct 22, 2025

Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics

James ReidOct 22, 2025

Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics