 Hello and welcome to the session. In this session we are going to discuss general theorems on differentiation. The first theorem states that differentiation of a constant is 0 that is dc by ds is equal to 0 where c is a constant. Let y be equal to c where c is a constant and mark the equation y is equal to c as 1. Let delta x be a small increment in x corresponding to this increment in x. Let there be an increment delta y in y then y plus delta y is equal to c mark this equation as 2. Now subtract equation 1 from equation 2 we get y plus delta y minus y which is equal to delta y is equal to c minus c that is 0. Now dividing both the sides by delta x we get delta y by delta x is equal to 0. Now taking limits as delta x tends to 0 we have limit delta y by delta x as delta x tends to 0 is equal to limit 0 as delta x tends to 0. Therefore we have limit delta y upon delta x as delta x tends to 0 can be written as dy by ds and is equal to limit 0 as delta x tends to 0 which is equal to 0. Therefore dy dx of y that is equal to c is equal to 0 for example dy dx of 8 which is a constant is equal to 0 and dy dx of 3 pi by 2 is also 0 that is when we differentiate a constant with respect to x we get 0. The second theorem states that an arbitrary constant disappears in differentiation that is dy dx of f of x plus c is equal to dy dx of f of x where c is constant let y be equal to f of x plus c and mark this equation as 1. Let delta x be a small increment in x and corresponding to this increment let delta y be increment in the value of y then y plus delta y is equal to f of x plus delta x plus c mark this equation as 2. As the trapped equation 1 from equation 2 we get y plus delta y minus y which is equal to delta y and is equal to f of x plus delta x minus of f of x. Now dividing both the sides by delta x we get delta y by delta x is equal to f of x plus delta x minus f of x all upon delta x. Now taking the limits as delta x tends to 0 we have limit delta y by delta x as delta x tends to 0 is equal to limit f of x plus delta x minus of f of x all upon delta x as delta x tends to 0. And therefore we have limit delta y upon delta x of delta x tends to 0 which can be written as dy by dx is equal to limit f of x plus delta x minus f of x upon delta x all delta x tends to 0 which can be written as f dash of x. Therefore we have dy by dx which is equal to dy dx of y that is f of x plus c is equal to f dash of x that can be written as dy dx of f of x. For example we have dy dx of x raised to the power 9 plus 5 which is equal to dy dx of x raised to the power 9 and is equal to 9 into x raised to the power 8. Therefore the value of dy dx of x raised to the power 9 plus 5 is equal to 9 into x raised to the power 8. The next theorem states that the differential coefficient of the product of a constant and a function is equal to the product of the constant and the differential coefficient of the function that is dy dx of c into f of x is equal to c into dy dx of f of x. Let y be equal to c into f of x be a function of x and mark this equation as 1. Let delta x be a small increment in the value of x corresponding to this let delta y be a small increment in the value of y then y plus delta y is equal to c into f of x plus delta x. Mark this equation as 2. Now subtract equation 1 from equation 2 we get delta y is equal to c into f of x plus delta x minus of f of x. Now dividing both the sides by delta x we get delta y by delta x is equal to c into f of x plus delta x minus of f of x whole upon delta x taking limits as delta x tends to 0 limit delta y by delta x. As delta x tends to 0 is equal to limit c into f of x plus delta x minus f of x whole upon delta x as delta x tends to 0 and therefore limit delta y by delta x as delta x tends to 0 can be written as dy by dx and is equal to As we know that limit of c into f of x can be written as c into limit of f of x therefore limit of c into f of x plus delta x minus of f of x upon delta x as delta x tends to 0 can be written as c into limit f of x plus delta x minus f of x whole upon delta x as delta x tends to 0. Therefore we have dy by dx is equal to c into limit f of x plus delta x minus f of x upon delta x as delta x tends to 0 is equal to dy dx of f of x. Therefore dy by dx that is dy dx of y which is equal to c into f of x is equal to c into dy dx of f of x for example if y is equal to 4 into x square then find the value of dy by dx. dy by dx can be written as dy dx of 4x square that is equal to 4 into dy dx of x square and therefore we have 4 into Now differentiating x square with respect to x we get 2 into x which is equal to 4 into 2 into x that is 8 into x therefore the value of dy by dx is equal to 8x. The next theorem states that the differential coefficient of the sum of two functions is equal to the sum of the differential coefficients of these functions that is dy dx of u plus v is equal to dy dx of u plus dy dx of v where u and v are functions of x whose differential coefficients exist. Let y be equal to u plus v and let u be equal to f of x and v be equal to d of x as u and v are functions whose differential coefficients exist. Therefore f of x and d of x are differential functions therefore dy dx of f of x is equal to limit f of x plus delta x minus f of x whole upon delta x delta x tends to 0 and d by dx of g of x is equal to limit g of x plus delta x minus of g of x whole upon delta x delta x tends to 0 Now d by dx of plus g of x that is dy by dx is equal to limit plus delta x plus g of x plus delta x minus of f of x plus g of x whole upon delta x as delta x tends to 0 We can also write it as dy by dx is equal to limit f of x plus delta x minus of f of x plus g of x plus delta x minus of g of x upon delta x as delta x tends to 0 That is we have g y by dx is equal to limit f of x plus delta x minus of f of x whole upon delta x as delta x tends to 0 plus limit g of x plus delta x minus of g of x whole upon delta x as delta x tends to 0 therefore we have dy by dx is equal to limit f of x plus delta x minus of f of x whole upon delta x as delta x tends to 0 can be written as dy by dx of f of x plus limit g of x plus delta x minus g of x by delta x as delta x tends to 0 can be written as dy dx of g of x therefore dy by dx is equal to dy dx of u plus dy dx of v as u is equal to f of x and v is equal to g of x Similarly dy dx of u minus v can be written as dy dx of u minus dy dx of v in general We can write dy dx of u plus v plus w plus and so on is equal to dy by dx plus dy by dx plus dy by dx plus and so on For example if y is equal to x cubed plus 2x square plus x minus 2 then find the value of dy by dx We have dy by dx which is equal to dy dx of y that is x cubed plus 2x square plus x minus 2 which can be written as dy dx of x cubed plus dy dx of 2x square plus dy dx of x minus dy dx of 2 and therefore we have on differentiating x cubed with respect to x we get 3x square plus on differentiating 2x square with respect to x we get 2 into 2x plus on differentiating x with respect to x we get 1 minus on differentiating 2 that is the constant with respect to x we get 0 Therefore we have 3x square plus 4x plus 1 therefore the value of dy by dx is equal to 3x square plus 4x plus 1 This is the detailed session hope you enjoyed this session