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Class XIName of the topic: CalculusWeek 4: 27th April to 3rd May, 2020 +Day 1Step – IFunction: If the value of a quantity y depends on the value of another quantity x, then y is the function of x. i.e y = f (x) The quantity y is called dependent variable and the quantity y is called independent variable. E.g y = 2x2 + 5x + 3 is a function of x.Differential co-efficient or derivative of a function:Let y = f(x) ------------------- (1)Let Δx be a small increment ( change) in x, and Δy be the corresponding small increment in y. Then y + Δy = f (x + Δx) -------------------- (2)(2) – (1) => Δy = f (x + Δx) – f (x) => Δy / Δx = f (x + Δx) – f (x) / ΔxHere Δy / Δx is called average rate of change of y with respect to x.Let us take Δx as small as possible i.e Δx 0, then Δy / Δx dy/dx.dy/dx is called differential co-efficient or derivative of y with respect to x.Theorems of differentiations:If y = c, where c is a constant, then dydx = 0If y = xn, where n is an integer, then dydx = n xn-1If y = c u, where u is a function of x, then dydx = c dudx.If y = u ± v ± w, then dydx = dudx ± dvdx ± dwdxIf y = u v, where u and v are functions of x, then dydx = u dvdx + v dudxIf y = u/v, then dydx = v dudx - u dvdx v2If y = un, then dydx = n un-1 dudx d dx (sin x) = cos x ddx ( sin u) = cos u dudx ddx ( cos x) = - sin x ddx ( cos u) = - sin u dudx ddx ( tan x ) = sec2 x ddx ( log x) = 1/x ddx ( log u) = 1/u dudx ddx ( ex) = ex ddx ( eu) = eu dudxChain rule: If y is a function of u and u is a function of x, then dydx = dydu . dudxStep – IICopy the above formulae in your C/W copy:End of Day – 1*Day 2Step – IGraphical interpretation of differentiation of a function: Graphically, the derivative of a function y = f(x) represents the slope of the tangent drawn to the function at a point.Second order differentiation: It is the rate of change of the rate of change of the function y with respect to x i.e d2ydx2 = ddx( dydx)Step – IISolve the following questions in your C/W copy:Differentiate the following with respect to x:4x3 + 7x2 + 6x + 9x5/2 – 5/x21/ √x1 / (x+1)( x2 + 3) ( x4 – 8)( x2 + x + 1) / ( x2 – x – 1) ( 3 + x2)1/2 √x - 1√xFind dydx :y = sin 2xy = x sin xy = cos 4xy = sin (ax + b)y = ( log x )2y = log ( ax + b)y = enxy = u3 + 2u and u = x + 1y = a sin θ and x = b cos θy = at2 and x = atIf s = 2t3 – 3t2 + 2, find the position, velocity and acceleration of a particle at the end of 2 second, s is measured in metre and t in second. End of Day – 2Day 3Step – IIntegral calculus or integration: It is an inverse process of differentiation. It is the process of finding the function whose derivative is given.Let ddx f(x) = F(x)Now, if we are given the derivative F(x) and we have to find the function f(x0 then this can be done with the help of Integral calculus.Definition: If the derivative of a function f(x) is F(x) then f(x0 is called the integral of F(x) w.r.t x. The integration of a function is written as –Fxdx = f(x)Formulae of integration:xn dx = xn+1n+1 + c, where c = constant. Here n ≠ -1dx = x + cx-1dx = log x + cexdx = ex + ce axdx = eaxa + cc u dx = c u dx, c = constantsinx dx = - cos x + ccosxdx= sin x + csinaxdx = -cosaxa + ccosaxdx = sinaxa + c( u ±v ±w)dx = u dx ± v dx ± w dxIntegration by parts:u v dx = u vdx – [ ddxu v dx] dxStep – IIWrite the above formulae in your CW copy.Integrate the following w.r.t x –x5x2 + 1xe3x( x - 1x )21√x4 e5xSin 5x + cos 4x7/x + 4√x – x4 + 10x cos xx e4xEnd of Day – 3Day – 4Step – IDefinite integral: the integral of a function whose minimum and maximum values are given is called definite integral.Suppose , we have a function f(x) whose minimum value is ‘a’ (lower limit) and maximum value ‘b’ (upper limit), then the integral of this function is written as abfxdx.In this case, first of all the given function is integrated using the appropriate formula and then the upper and lower limits of the function are substituted. abfxdx = F(b) – F(a)Step – IISolve the following in your C/W copy.12x3dx–π2π2sinx dxuvmv dv0π/2cosx dx∞RGMmx2dx End of Day – 4Day – 5Solve the following in your C/W copy:Differentiate with respect to x;sin3 (4x + 1) ii) cos (ax2 + b)iii) e-x cos x iv) log (1x)v) sin ( 1 – 2x)2 If x = a (cos θ + θ sin θ) and y = (sin θ – θ cos θ), find dydx .The area A of a blot of ink is growing such that after t second A = 3t2 + t5 + 7. Calculate the rate of increase of area after 5 second.Integrate the following with respect to x:( √x - 1√x )2 ii) x2 + sin x iii) ( ax + b)n iv) log x v) 10 x4 - √7 + 2√x 5. Solve the following – i) 0π/4sin4x dx ii) 0Qqc dq, where c is constant iii) 04√x dx iv) 01( 2x2+ 5x-12) dxEnd of Day – 5 ................
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