CALCULUS

🎬2020371h 51mNigeriaMath

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Description

The TV series CALCULUS, created by an unlisted creator and produced by an unlisted studio, first premiered on March 21, 2020 in Nigeria. The series spans 1 season with 50 episodes, featuring the voices of an unlisted voice cast. It follows the story of 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:. This is a Math series and has received a rating of 0/10 from 0 viewers.

Where to Watch

ugc-kid-edu.comPrince Nelson Enwerem

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Filming Location

127A Smithfield Road, Frederiksted, Virgin Islands

Production

Castle Rock Entertainment

Award

21 wins & 43 nominations total

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User Review

IMVU_jxt_•22/10/25 07:46

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.

Ewurafua22/10/25 07:46

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 Rima22/10/25 07:46

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 22/10/25 07:46

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)]

provoicelameck22/10/25 07:46

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🇮🇳🇲🇼22/10/25 07:46

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❤️‍🔥22/10/25 07:46

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)]

THECUTEABIOLA22/10/25 07:46

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 Rokaya22/10/25 07:46

Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics

James Reid22/10/25 07:46

Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics

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