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education/math/MATH1210 (calc 1)/Derivatives.md
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@ -110,7 +110,13 @@ for all $a > 0$
# Chain Rule # Chain Rule
$$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$ $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
## Examples
> 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 $g(x) = x^2 + 3)$.
1. First find the derivative of the outside function function ($f(x) = x^4$):
$$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 $$
2.
# Trig Functions # Trig Functions
$$ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 $$ $$ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 $$
$$ \lim_{x \to 0} \dfrac{\cos x - 1}{x} = 0 $$ $$ \lim_{x \to 0} \dfrac{\cos x - 1}{x} = 0 $$
@ -143,8 +149,8 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
> Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$ > Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$
1. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$ 3. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
2. $= 4x^\frac{1}{3} - x^{-6}$ 4. $= 4x^\frac{1}{3} - x^{-6}$
3. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$ 5. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$
4. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$ 6. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
5. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$ 7. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$