From f6bf13f9e09229878a303fcfe3bdcb675962b9ed Mon Sep 17 00:00:00 2001 From: arc Date: Tue, 18 Feb 2025 10:06:11 -0700 Subject: [PATCH] vault backup: 2025-02-18 10:06:11 --- .../math/MATH1210 (calc 1)/Derivatives.md | 26 ++++++++++++------- 1 file changed, 16 insertions(+), 10 deletions(-) diff --git a/education/math/MATH1210 (calc 1)/Derivatives.md b/education/math/MATH1210 (calc 1)/Derivatives.md index 4e517c8..f4c1b3a 100644 --- a/education/math/MATH1210 (calc 1)/Derivatives.md +++ b/education/math/MATH1210 (calc 1)/Derivatives.md @@ -148,22 +148,28 @@ This is used when you want to take the derivative of a function raised to a func 9. $\dfrac{dy}{dx} = (\ln(x) + 1)x^x$ ### Examples +> Find the derivative of $(7x + 2)^x$ + +1. $\ln y = \ln((7x+2)^x)$ +2. $\ln y = x*\ln(7x + 2)$ +3. $\dfrac{dy}{dx} \dfrac{1}{y} = 3x*\dfrac{1}{7x + 2} * 3\ln(7x+2)$ + > Find the derivative of the function $y = (2x \sin x)^{3x}$ -1. $\ln y = \ln (3x \sin x)^{3x}$ -2. $\ln y = 3x * ln(2x \sin x)$* -3. $\dfrac{d}{dx} \ln(y) = \dfrac{d}{dx} 3x(\ln 2 + \ln x + \ln(sinx))$ -4. $\dfrac{1}{y} \dfrac{dy}{dx} = 3(\ln 2 + \ln x + \ln(\sin(x))) + 3x (0 + \dfrac{1}{x} + \dfrac{1}{\sin x} * \cos x)$j -5. $\dfrac{dy}{dx} = (3\ln 2 + 3 \ln x + 3\ln \sin(x) + 3\ln(\sin(x) + 3x\cot(x))(2x\sin x)^{3x}$ +4. $\ln y = \ln (3x \sin x)^{3x}$ +5. $\ln y = 3x * \ln(2x \sin x)$* +6. $\dfrac{d}{dx} \ln(y) = \dfrac{d}{dx} 3x(\ln 2 + \ln x + \ln(sinx))$ +7. $\dfrac{1}{y} \dfrac{dy}{dx} = 3(\ln 2 + \ln x + \ln(\sin(x))) + 3x (0 + \dfrac{1}{x} + \dfrac{1}{\sin x} * \cos x)$j +8. $\dfrac{dy}{dx} = (3\ln 2 + 3 \ln x + 3\ln \sin(x) + 3\ln(\sin(x) + 3x\cot(x))(2x\sin x)^{3x}$ # Chain Rule $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$ ## 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 $g(x) = x^2 + 3)$. -6. First find the derivative of the outside function function ($f(x) = x^4$): +9. First find the derivative of the outside function function ($f(x) = x^4$): $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$ -7. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. +10. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. $$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$ > Apply the chain rule to $x^4$ @@ -199,7 +205,7 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$ - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$. Given these facts: -8. Let $y$ be some function of $x$ -9. $\dfrac{d}{dx} x = 1$ -10. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\ +11. Let $y$ be some function of $x$ +12. $\dfrac{d}{dx} x = 1$ +13. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\