What is the purpose of Differentiation and Integration, in Calculus?
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There are many reasons why one would "want" to.\nIf you mean: 'as a layman; what is the practical point of this' then (basically) differentiation is a way of finding a tangent and integration is a way of finding "the area under the graph".\nAs a Mathematician, my motives are somewhat different; I consider Differentiation and Integration as interesting functions on function spaces. For example, Integration of Real-valued functions of on Reals is really of interest as a "measure" of various sets. The topic of Measure Theory is a large and interesting one.\n\nBut for your purposes I guess it suffices to say "because you get useful information about the function".
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Si? Are you referring to the sine integral? If so, you need only remember two things:\nDifferentiation is the inverse of integration\nd/dx f(g(x)) = f'(g(x)) * g'(x)\n\nSo:\n\nd/dx (Si(x²)) = sin (x²)/(x²) * 2x
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This is a Frequently Asked Question. Following is a modified previous answer:\n\n\nIf one has a function f as a map from the real line into itself, one also has the graph of the function as the set of all points (x,f(x)) as a curve in the Cartesian plane. If one wanted to know the area between the x-axis and this curve over some interval on the x-axis, one would use a definite integral. If one wanted to know the slope of a line tangent to the curve at some point (x,f(x)) on the curve, one would use the derivative of the function f at the point x. Both of these operators are very useful in physics, as most physical properties are defined as changes in some property. If p(x) is a graph of some physical property over the domain it is a function of, then the change in p with respect to x approaches the slope of the tangent line to p(x) as one considers smaller and smaller changes in x. Thus, the derivative is the ideal way of talking about continuous changes in physical properties.\n\nFor differentiation in one dimension, you can use the definition: the derivative of a function f at the (domain) point x is the limit as h approaches 0 of the slope of the secant line [f(x+h)-f(x)]/h. Visually, we take a secant line to the curve that passes through the points (x, f(x)) and (x+h, f(x+h)) and rotate it (see what happens as h gets smaller) until it only touches the curve once (it is now tangent to the curve). This line is unique at each point.\nIn more than one dimension, using the definition is cumbersome for anything other than proving when a derivative does or does not exist, so we use a useful theorem that says the derivative of a function f from R (the real line) into R^n (n-dimensional Euclidean space) is just the vector of partial derivatives of f. Ie., if f(x,y) = x^2 + y^2 (the graph is a circular paraboloid), then Df(x,y) = (df/dx, df/dy) = (2x, 2y). Partial derivatives are taken relative to one variable only; they end up treating the other variables as constants. The technical definition of the partial of f with respect to x at the point x0 is what you suspect it would be: the limit as h approaches zero of the slope of the secant line [f(x0+h, y) - f(x0,y)]/h.\nOnce one gets to these derivatives of multivariable functions though, one must be very careful, as the existence of all partial derivatives at a point does not mean the derivative exists at that point.\nOnce one gets a bit of experience taking derivatives using the definition, one can derive some simple derivative rules which allow you to take derivatives of elementary functions without explicitly doing all that limit mumbo-jumbo. A list of common differentiation rules can be found here: http://mathworld.wolfram.com/Derivative.html .\n\nIntegration is a more delicate matter. If you're doing a definite Riemann integral (finding the area under a curve), it is the limit of a Riemann sum as you increase the number of rectangles under the curve ( See http://mathworld.wolfram.com/RiemannSum.html ). The indefinite integral is just the function that will return the definite integral when you take the difference between two points.\nSometimes the rectangles will not converge to any finite value or it does not have a unique convergence and we say the function is not integrable on that interval (Take f(x) = 1/x on the interval [0,1] for example). A more technical definition of integration will help integrate more pathological functions, but you can look up this definition yourself in a real analysis text; it's not that different from summing rectangles.\nAgain, with integrals we have a very nice theorem that tells us that the indefinite integral of the derivative of a function f is f itself. This is the fundamental theorem of calculus, and it takes a lot of work out of finding an integral. For example, if f(x) = 2x, we know from our experience with taking derivatives that the indefinite integral of f(x) is just F(x) = x^2 + C where C is an unknown constant since the deriv
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Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
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A complex function is differentiable at a point z0 if the limit z -> z0  (f(z) - f(z0))/(z - z0) exists. However, the differivative no longer has any geometrical interpretation.\n\nThe differentiation can be done in a fashion that resembles real differentiation using the definition:\n\nlim dz -> z   ( f(z0 + dz) - f(z0) ) / dz\n\nHowever since z is a complex plane, the limit as dz -> z can approach from any direction. If it is differentiable, the value of the derivative is the same when approached from any direction, so a path is chosen that makes the calculations easy, typically keeping the real or imaginary part constant and just approaching in a direction parallel to either axis. The effect is that the limit calculation is done on a single variable. But this only finds the derivative at a single point.\n\nEquations for the derivative of a function can be found by using the Cauchy-Riemann Equations, where valid, if \nw = f(z) = u(x,y) + i v(x,y)\n\nin which case if partial du/dx = dv/dy AND partial dv/dx = -du/dy AND all theses derivatives are continuous, then \n\ndw/dz = partial du/dx + i dv/dx = partial dv/dy - i du/dy
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well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
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in calculus, integration is just an extension of sum. for discrete values, we use the summation. for continuous values, we use integration.\n\nindefinite integration is just integrating indefinitely, ie., without any bounds. this usually results in a polynomial or trigonometric equation etc. \n\ndefinite integration is done between two specific points, be it on the x axis or y axis. this results in a constant value, rather than a polynomial or other equations.\nthe applications of definite integration is to find the area under curves, etc..
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The best ? Go for Richard Courant's "Differential and Integral Calculus". Every structure is motivated and all of the exercises are enlightening instead of busy work.\nIf you have the maturity, Apostol's "Calculus" is another classic introduction and a good companion.
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differentiation is the process of finding the rate of change of a gradient.\n( ^= to the power)\n\nsay you are given an equation y=x^n\ndo differentiate, take the power of x, and multiply it by its coefficient. in this case, x has a coefficient of 1. \ny=nx^n\nthe next step it to subtract 1 from the base of x (its power)\ny=nx^n-1\ndifferentation is now complete\n\ne.g.\ny=3x^4\ndy=12x^3\nno matter what the power of x is, just multiply it by the coefficient of x, and subtract 1 from it. (do not subtract then multiply)\n\nIntergration is the opposite, and it too complicated fot this post. talk to you teacher, of go to www.s-cool.co.uk
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The integration by parts formula is\nint(u*dv)=u*v-int(v*du)\n\nLet u=r and dv=e^(r/2)dr\n\nNow take the derivative of u with respect to r:\ndu=dr\n\nNow integrate dv\nint(dv)=int[e^(r/2)dr]\nv=[e^(r/2)]/(1/2)  --> The chain rule was used here. \n__=2e^(r/2)\nWhen you take the derivative you multiply by "what's inside" so when you integrate, you have to divide by it.\n\nNow plug into the formula\n2re^(r/2)-int(2e^(r/2)dr)\n\n=2re^(r/2)-2(e^(r/2))/(1/2)+C\n\n=2re^(r/2)-4(e^(r/2))+C\n\n=(r-2)*2e^(r/2)+C
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Try the inside cover of your Calculus book - it usually summarizes the rules for derivatives and some key integrals.
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Calculus was confusing enough in English - I can't imagine taking it in Japanese.  You might want to try to get hold of a textbook or something in English.\n\nIt's an integral.  More specifically an indefinite integral.  It's kind of like the inverse of the derivative function except for that there is no way to get the constant back (hence the "C").  Basically the indefinite integral is the set of all functions whose derivative is equal to the thing between the integral sign and the dx (or dy, dq, whatever the variable is... I'm blanking on general name of the thing).  The integral symbol tells you it's an integral (obviously), and the "x" in dx tells you what variable you're integrating with respect to (that will make more sense when you learn to integrate more complicated expressions).  So the indefinite integral of x^2 is 1/3x^3 + C with respect to x where C is some constant because the derivative of 1/3x^3 + C is x^2 (remember that the derivative of a constant is zero).\n\n***A Note on Definite and Indefinite Integrals***\nA couple of the answers below appear to be describing the definite integral.  The definite integral can be described as the sum of the areas of rectangles, and can be used to find the area under a curve (among other things).  The indefinite integral (or antiderivative), however, which is the one refered to in the question, describes a function.  Although the function described by the indefinite integral is extremely useful in calculating the area under a curve, it does not directly describe this area.  In addition to being used to caculate the definite integral, the indifinite integral can also be used, for example, to find an equation for the position of an object given an equation for its velocity.\n\nAnother afterthough: I may have gotten some of the terminology wrong as my high school math department had a strange affinity for inventing their own terms.
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Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
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In basic calculus, we use the derivative to find the slope of a curve, and the integral to find the area under a curve.  The question is, why are these operations the opposite of each other?\n\nThe quick answer is:  The derivative is used to calculate the rate of change of a function.  And the rate of change of the area under a curve is actually the original function.\n\n\nLet's say you have a simple curve, y=f(x).\n\nThe derivative of the function is gives you the rate of change of that function.  As x increases by 1, how much does y change?  That's also known as the slope.\n\n[And you probably know that the derivative of the derivative is also useful.  That's the rate of change of the slope.  When the slope stops changing (or its rate of change=0), it's at a maximum, minimum, or point of inflection.]\n\nThe same relationship exists for the area under a curve and the function itself.  You probably learned to think of the integral of f(x) as the sum of the thin vertical strips under the curve.  That's the area under the curve.  As you move to the right, how fast does the area grow?  You're essentially adding these vertical strips to the area.  The length of a vertical strip is f(x), or the original function.  So the original function f(x) represents how fast the area grows for any value x.\n\n\nETA: What do I have to do to get selected as best answer?
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Calculus is important because it is the branch of mathematics that allows you to examine more than one variable at once.  Time and speed or speed and position for instance.  It can also get more complicated as you add more variables.  Because you live in a world with many variables it is important to have a type of mathematics that deals with many variables.  \nThis is why it has applications in Business, Economics, Physics, Engineering and just about anything else you can think of.
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Differentiating sinx gives cos x.\n\nDifferentiating sin[f(x)] gives f'(x)cos[f(x)]\n(The chain rule)\n\nf(x) = 5x\n\nso f'(x) = 5\n\nHence:\n\n5cos(5x)
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For all the rules of differentiation and integration, check sources