dy/dx = 3 integral = (3x^2)/2
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
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
y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)
y = 4x2 - 2 is a parabolic function with a focal point at the location (0, -2). It's derivative can be expressed as: dy/dx = 8x It's indefinite integral can be expressed as: ∫(4x2 - 2) dx = (4x3)/3 - 2x + C
dy/dx = 3 integral = (3x^2)/2
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
∫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
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.
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
You have : y = e^(5x)^2 and de^u/dx = [ e^u ] [ du/dx ]dy/dx = [ e^(5x)^2 ] [ 10x ]
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)
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
You take the change in Y or dy and divide it by the change in X or dx. Slope equals dy/dx.
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.
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.
by implicit differentiation you have y+x*dy/dx-2dy/dx=0 solving for dy/dx you'll have dy/dx=y/(2-x) and solving for y in the original equation and plugging it back in, you'll get dy/dx=1/(-x^2 +4x-4) which is your final answer