29 lines
2.0 KiB
Markdown
29 lines
2.0 KiB
Markdown
# Antiderivatives
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An antiderivative is useful when you know the rate of change, and you want to find a point from that rate of change
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> A function $F$ is said to be an *antiderivative* of $f$ if $F'(x) = f(x)$
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## Examples
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> Find the antiderivative of the function $y = x^2$
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1. We know that $f'(x) = 2x^1$
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## Formulas
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| Differentiation Formula | Integration Formula |
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| ----------------------------------------------------- | ------------------------------------------------------- |
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| $\dfrac{d}{dx} x^n = nx^{x-1}$ | $\int x^n dx = \dfrac{1}{n+1}x^{n+1}+ C$ for $n \ne -1$ |
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| $\dfrac{d}{dx} kx = k$ | $\int k \space dx = kx + C$ |
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| $\dfrac{d}{dx} \ln \|x\| = \dfrac{1}{x}$ | <br>$\int \dfrac{1}{x}dx = \ln \|x\| + C$ |
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| $\dfrac{d}{dx} e^x = e^x$ | <br>$\int e^x dx = e^x + C$ |
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| $\dfrac{d}{dx} a^x = (\ln{a}) a^x$ | $\int a^xdx = \ln \|x\| + C$ |
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| $\dfrac{d}{dx} \sin x = \cos x$ | $\int |
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| $\dfrac{d}{dx} \cos x = -\sin x$ | |
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| $\dfrac{d}{dx} \tan{x} = \sec^2 x$ | |
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| $\dfrac{d}{dx} \sec x = \sec x \tan x$ | |
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| $\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$ | |
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| $\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}$ | |
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| $\dfrac{d}{dx} k f(x) = k f'(x)$ | |
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| $\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)$ | |
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