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vault backup: 2025-02-16 19:02:21

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27
.obsidian/plugins/obsidian-git/data.json vendored
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@ -1,27 +0,0 @@
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31
education/math/MATH1210 (calc 1)/Derivatives.md
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@ -15,6 +15,15 @@ If we have the coordinate pair $(a, f(a))$, and the value $h$ is the distance be
$$\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}$$ $$\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}$$
The above formula can be used to find the *derivative*. This may also be referred to as the *instantaneous velocity*, or the *instantaneous rate of change*. The above formula can be used to find the *derivative*. This may also be referred to as the *instantaneous velocity*, or the *instantaneous rate of change*.
## Examples
> Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$
1. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
2. $= 4x^\frac{1}{3} - x^{-6}$
3. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$
4. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
5. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
# Point Slope Formula (Review) # Point Slope Formula (Review)
$$ y - y_1 = m(x-x_1) $$ $$ y - y_1 = m(x-x_1) $$
Given that $m = f'(a)$ and that $(x_1, y_1) = (a, f(a))$, you get the equation: Given that $m = f'(a)$ and that $(x_1, y_1) = (a, f(a))$, you get the equation:
@ -114,9 +123,9 @@ $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
> Given the function $(x^2+3)^4$, find the derivative. > Given the function $(x^2+3)^4$, find the derivative.
Using the chain rule, the above function might be described as $f(g(x))$, where $f(x) = x^4$, and $g(x) = x^2 + 3)$. Using the chain rule, the above function might be described as $f(g(x))$, where $f(x) = x^4$, and $g(x) = x^2 + 3)$.
1. First find the derivative of the outside function function ($f(x) = x^4$): 6. First find the derivative of the outside function function ($f(x) = x^4$):
$$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$ $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
2. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 7. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$.
$$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$ $$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$
> Apply the chain rule to $x^4$ > Apply the chain rule to $x^4$
@ -152,18 +161,8 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
- Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$. - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$.
Given these facts: Given these facts:
1. Let $y$ be some function of $x$ 8. Let $y$ be some function of $x$
2. $\dfrac{d}{dx} x = 1$ 9. $\dfrac{d}{dx} x = 1$
3. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\ 10. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
What's the derivative of $y^2$?
$\dfrac{d}{dx} y^2 = 2(y)^1 *\dfrac{dy}{dx}$
# Examples ## Examples
> Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$
4. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
5. $= 4x^\frac{1}{3} - x^{-6}$
6. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$
7. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
8. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$