# Calculus

## The branch of mathematics that deals with the study of continuously changing quantities, with the use of limits and the differentiation and integration of functions of one or more variables, is called Calculus. Calculus analyzes aspects of change in processes or systems that can be modeled by functions. The English physicist, Isaac Newton, and the German mathematician, G. W. Leibniz, working independently, developed calculus during the 17th century.

### How do you differentiate 2tanx? Coefficients can be removed when differentiating, i.e. d/dx 2tanx = 2 d/dx tanx You should know that d/dx tanx = sec2x (either by differentiating sinx/cosx or by just remembering the derivatives of common trig functions, as it will come in handy). So the answer is 2sec2x

### What is the indefinite integral of 2x ln2x dx? In order to work out this problem, we need to learn how to apply the integration method correctly. The given expression is ∫ 2xln(2x) dx. Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx or du/2 = dx. This method is both valid and easy to avoid working out with too much expressions. You should get: ∫ uln(u) (du/2) = ½ ∫ uln(u) du Use integration by parts, which states that: ∫ f(dg) = fg - ∫ g(df) We let: f = ln(u). Then, df = 1/u du dg = u du. Then, g = ∫ u du = ½u² Using these substitutions, we now have: ½(½u²ln(u) - ½∫ u du) = ¼(u²ln(u) - ∫ u du) Finally, by integration, we obtain: ¼ * (u²ln(u) - ½u²) + c = 1/8 * (2u²ln(u) - u²) + c = 1/8 * (2(2x)²ln(u) - (2x)²) + c = 1/8 * (2x)² * (2ln(u) - 1) + c = ½ * x² * (2ln(2x - 1)) + c

### How do you solve x equals 2 sin x? The problem x = 2 sin x cannot be solved by using algebraic methods. One solution is to draw the graphs of y = x and y = 2 sin x. The two lines will intersect. The values of x where the intersection takes place are the solutions to this problem. You can tell from the graph that one solution is x=0 and verify this contention by noting that 2 sin(0) = 0. You can find the other solution through numerical methods and there are many that will give you the correct solution. Perhaps the simplest is to start with a value of X like pi/2 and then take the average of 2*sin(X) and X. Using that as your new value, again take the average of 2*sin(X) and X. As you continue to do this, the value will get closer and closer to the desired value. After 20 steps or so, the precision of your calculator will probably be reached and you will have a pretty good answer of about 1.89549426703398. (A spreadsheet can be used to make these calculations pretty easily.)

### What is the anti-derivative of 5x? ∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1. Apply that rule to get: ∫ 5x dx = 5 ∫ x dx [Factor out the constant] = 5 ∫ x1 dx [Make note of the exponent for x] = 5x1 + 1/(1 + 1) + c = (5/2)x2 + c

### What is the antiderivative of -cscxcotx? First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.

### If f of x equals 10 plus x squared then f of x and h equals? If f(x)=x2+10, then f(x+h)=? f(x+h)=(x+h)2+10 (since f(x)=x2+10, substitute the x in x2 to (x+h)2)=(x+h)(x+h)+10 (then multiply (x+h) by (x+h) by doing the FOIL method)=x2+xh+xh+h2+10 (First: x*x, Outside: x*h, Inside: h*x, Last: h*h)=x2+2xh+h2+10 (combine like terms (xh+xh=2xh)) So if f(x)=x2+10, then f(x+h)=x2+2xh+h2+10

### What is the integral of 3x? In order to compute that integral, we need to use the power rule: ∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1. Apply that rule to get: ∫ 3x dx = 3 ∫ x dx [Factor out the constant] = 3 ∫ x1 dx [Make note of the exponent] = 3x1 + 1/(1 + 1) + c = 3x2/2 + c So that is the integral of 3x.

### Can the first quartile equal the third quartile? yes, if all the data is the same number; when the range is zero. * * * * * That is not true. You need 25% of the values to be small, then 50% identical values, followed by 25% large values. Then the lower (first) quartile will be the same as the upper (third) quartile. The inter-quartile range (IQR) will be zero but the overall range can be as large as you like.

### What is the derivative of 2Y? It depends: The derivative with respect to Y is: dX/dY 2Y = 2 The derivative with respect to X is: dY/dX 2Y = 0

### What are some real life applications of calculus? Pretty much any serious statistical model or experiment on anything will use basic calculus to interpret data. Anything that exponentially grows or decays (radioactive matter, bacteria, population growth, etc.) Anything that's built to be structurally sound. Anything that uses the EM spectra (radio, microwaves, visible light, etc.) All scientific industries use calculus practically constantly. And on and on and on... In reality, it's rarely pure theoretical calculus that's being done. Rather, another branch of math based on and built from the principles and results of calculus is primarily used called differential equations. Don't forget integration, the other "half" of calculus. That is as equally important in your listed applications. Also, both theoretical and applied calculus use both differentiation and integration. Differentiation isn't a separate branch of maths, but one of the two major branches of calculus as a whole.

### What is the integral of -e to the -x? Know that ∫eu du = eu du/dx + c ∫-e-x dx = e-x + c But ∫eu du = eu + c Perhaps we are integrating -(e-x ) though the question might be (-e)-x Question is not clear.

### What is the geometrical interpretation of line integral? ### What is the derivative of 2 over X? f(x) = 2/x = 2x^-1 so dy/dx = -2x^-2 = -2/x² f(x) = 2/x This can be re-written as f(x) = 2x^-1 according the to the laws of exponents. Now, we just take the derivative normally: f(x) = 2x^-1 f ' (x) = -1 × 2x^(-1-1) = -2x^-2 The above can be re-written in quotient form as f ' (x) = -2/x²

### How do you integrate periodic functions? Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.

### What is retardation in physics? Retardation is the application of a force that produces negative accelleration. Synonyms would be braking, decelleration, damping, etc. Gravitational force operates downward (in a negative direction) so, in most frames of reference, gravity is a retarding force.

### What is the integral of 1 divided by x3 plus x? Note that (1/x³) + x = x-3 + x Then, by the power rule, we integrate the expression to get: x-3 + 1/(-3 + 1) + x1 + 1/(1 + 1) + c = -x-2/2 + x2/2 + c where c is the arbitrary constant

### What is the derivative of 2cosx times the inverse of sinx? d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).

### Formula for velocity? V = d / t (or v = dd/dt & v = integral of acceleration with respect to time for physics involving calculus) where V is velocity, d is displacement (distance traveled) and t is time. vf = vi + a*t (vf)^2 = (vi)^2 + 2*a*d Where vf is final velocity, vi is initial velocity, and a is acceleration v = p/m Where p is object's momentum and m is object's mass  