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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
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 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.
Why is ddx the derivative operator for Calculus?
It is the operator used by Gottfried Liebniz, who developed calculus (in parallel with Newton).The "d" denotes a difference and, the full representation, dy/dx (not ddx), is used to represent a change in y divided by (per) a change in x. However, calculus goes further than comparing simple changes but uses these changes for smaller and smaller changes in x, that is, the limit of the change as dx approaches 0.
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).
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 is the derivative of sinx0?
we have to find out dy/dx of sinx0 ----
Differentiate y equals ax?
dy/dx = a
