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We have:
int int (x * sin(y)) dx dy
Integrate x first:
int(x)dx = 1/2 * x2 + C
Now integrate sin(y):
int(sin(y))dy = -cos(y) + C
Multiply:
-1/2 * x2 * cos(y) + C
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How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
How do you differentiate tan2 x?
By using the chain rule: dy/dx = dy/du x du/dx With y = tan2x Let u = tan x Then: y = u2 du/dx = d/dx tan x = sec2x dy/dx = dy/du x du/dx = 2u sec2x = 2 tan x sec2x
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
If y equals x cube plus 2x and dx over dt equals 5 what is dy over dt when x equals 2?
y=x3+ 2x, dx/dt=5, x=2, dy/dt=? Differentiate the equation with respect to t. dy/dt=3x2*dx/dt Substitute in known values. dy/dt=3(2)2 * (5) dy/dt=60
Find the derivatives of y equals 2 sin 3x?
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
Integral of sin square root x?
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
What is the integral of x3 ex4 dx?
∫x3ex4 dx = 1/4ex4 + c To solve, let y = x4, then: dy = 4x3 dx ⇒ 1/4dy = x3 dx ⇒ ∫x3ex4 dx =∫ex4 x3 dx = ∫ey 1/4 dy = 1/4ey + c but y = x4, thus: = 1/4ex4 + c
How do you express an iterated integral in six different ways?
First, draw the region/solid being bounded by parameters say: y^2 + z^2 = 9, x = -2, and x = 2 Now analyze what possible iterated integrals can be used to find this region. the two "main" iterated integrals are: the triple integral from [-2,2] [-3,3] [-sqrt(9-y^2),sqrt(9-y^2)] dz dy dx and [-2,2] [-3,3] [-sqrt(9-z^2),sqrt(9-z^2)] dy dz dx Now, instead of sketching every region to find the different possible integrals, using the rules of triple integration, they will essentially be any legal alteration of the order of the "main" integrals. essentially, the first main integral can be rewritten as dx dz dy, and dz dx dy the second can be written as dx dy dz and dy dx dz.
How do you differentiate tan2 x?
By using the chain rule: dy/dx = dy/du x du/dx With y = tan2x Let u = tan x Then: y = u2 du/dx = d/dx tan x = sec2x dy/dx = dy/du x du/dx = 2u sec2x = 2 tan x sec2x
How do you find dy by dx of y equals e square 5x?
You have : y = e^(5x)^2 and de^u/dx = [ e^u ] [ du/dx ]dy/dx = [ e^(5x)^2 ] [ 10x ]
What is the integral of 2?
The integral of 2 is "who gives a $%&#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
How do you solve a slope?
You take the change in Y or dy and divide it by the change in X or dx. Slope equals dy/dx.
How do you differentiate Tan2x?
Chain Rule: let u=2x and y=tan(u) du/dx = 2 and dy/du = sec^2(u) dy/dx = du/dx x dy/du multiply them together and replace u=2x into the equation.. therefore dy/dx = 2(sec^2(2x)) hope that helps.
How do you prove the derivative of parametric equations?
The question is to PROVE that dy/dx = (dy/dt)/(dx/dt). This follows from the chain rule (without getting into any heavy formalism). We know x and y are functions of t. Given an appropriate curve (we can integrate piece-wise if necessary), y can be written as a function of x where x is a function of t, i.e., y = y(x(t)). By the chain rule, we have dy/dt = dy/dx * dx/dt. For points where the derivative of x with respect to t does not vanish, we therefore have (dy/dt)/(dx/dt) = dy/dx.
What is the derivative of xy-2y equals 1?
To find the derivative of the equation xy - 2y = 1, differentiate each term with respect to x using the product rule for the first term and the power rule for the second term. This yields dy/dx = (1 - 2x).
