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Calculus Concepts |
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Rules |
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Sum and Difference Rule: d/dx (u + v) = du/dx + dv/dx Example: d/dx (5x + x^2) u = 5x, so du/dx = 5 and v = x^2, so dv/dx = 2x Therefore the answer is 5 + 2x |
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Constant – Multiple Rule: d/dx (cu) = c du/dx Example: d/dx (3x) u = x, so du/dx = 1
and c = 3 because 3 is
the constant Therefore the
answer is 3 * 1 = 3 |
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Product Rule: d/dx (uv) = u dv/dx + v du/dx Example: d/dx [sin(x)][cos(x)] u = sin(x), so du/dx = cos(x) and v = cos(x), so dv/dx = -sin(x) Therefore the answer is [sin(x)][-sin(x)] + [cos(x)][cos(x)] or -[sin(x)]^2 + [cos(x)]^2 |
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Quotient
Rule: d/dx
(u/v) = [v(du/dx) – u(dv/dx)]/[v^2] Example: d/dx [sin(x)]/[cos(x)] u = sin(x), so du/dx = cos(x) and v = cos(x), so dv/dx = -sin(x) Therefore the answer is [[cos(x)][cos(x)] + [sin(x)][-sin(x)]] / [cos(x)]^2 or [[cos(x)]^2
– [sin(x)]^2] / [cos(x)]^2 |
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Power Function: d/dx u^n = [n][u^(n-1)] du/dx Example: d/dx
x^4 u
= x, so du/dx = 1 and n = 4 Therefore the
answer is [4][x^3] * 1 or 4 x^3 |
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Integration by Parts: The
integration of u dv = uv – the integration of v du Example: The integration of ln(x) dx u = ln(x), so du = (1/x) dv = 1 dx, so v = x Therefore the answer is [ln(x)][x] – the integration of x[1/x] or [x][ln(x)] – the integraton of
1 = [x][ln(x)] - x |
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Power Function: The
integration of u^n du = [1/(n+1)] u^(n+1) + C Example: The integration of x^3 u = x, so du = 1 and n = 3 Therefore the answer is (1/4) u^4 * 1 or (u^4)
/ 4 |
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Area Formulas: Circle: (pi)r^2 Triangle: (1/2)bh = (1/2)ab sin(c) Parallelogram = bh Trapezoid = (h/2)(b1 + b2) Volume Formulas: Right Circular Cone: (pi)r^2 h) / 3 Sphere: (4/3) (pi)r^3 |
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Rules |
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