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If y equals the integral from 5x on top to cosx on bottom of cos of u squared du what is y'?
y=S^5x _cos(x) cos(u²) du The derivative of a definite integral of a function f(x) is equal to the difference in the product of the function at each limit of integration times the limit of integration. y'=cos(u²)*du/dx from u=cos(x) to u=5x y'=-sin(x)*cos(cos(x)²)-5*cos(25x²) To understand why this works, consider the following where F(x) is the antiderivative of f(x) y=F(g(x))-F(h(x))=S f(x)dx from h(x) to g(x) If you take the derivative of this expression and apply the chain rule dy/dx = dF(g(x))/dx - dF(h(x))/dx = f(g(x))*dg/dx - f(h(x))*dh/dx
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.
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 derivative of the square root x?
Use: √x = x1/2 By the Power Rule (Decrease the power by 1. Multiply by the original power.): d/dx √x = d/dx x1/2 = 1/2 x-1/2
How do you find the delta x and delta y on a graph?
The delta x and delta y (dx and dy) are the changes in x and in y. When you take the derivative you are provided with a ratio (rise over run) of these changes. Take f(x)=3x^2 + 2x Derivative: (dy/dx)=6x + 2 From this, you can solve for dx and dy algebraically.
