vault backup: 2025-02-18 09:56:11

education/math/MATH1210 (calc 1)/Derivatives.md

@@ -139,19 +139,28 @@ This is used when you want to take the derivative of a function raised to a func
 
 1. $\dfrac{d}{dx} x^x$
 2. $y = x^x$
-3. $\ln y = \ln x^x$
+3. Take the natural log of both sides: $\ln y = \ln x^x$
 4. $\ln(y) = x*\ln(x)$
-5. $\dfrac{d}{dx} \ln y = \dfrac{d}{dx} x \ln x$
-6. $\dfrac{1}{y} \dfrac{dy}{dx} = 1 * \ln x + x * \dfrac{1}{x}$ 
+5. Use implicit differentiation: $\dfrac{d}{dx} \ln y = \dfrac{d}{dx} x \ln x$
+6. Solve for $\dfrac{dy}{dx}$: $\dfrac{1}{y} \dfrac{dy}{dx} = 1 * \ln x + x * \dfrac{1}{x}$ 
+7. $\dfrac{dy}{dx} = (\ln x + 1) * y$
+8. Referring back to step 2, $y = x^x$, so the final form is:
+9. $\dfrac{dy}{dx} = (\ln(x) + 1)x^x$
+
+### Examples
+> Find the derivative of the function $y = (2x \sin x)^{3x}$
+
+1. $\ln y = \ln (3x \sin x)^{3x}$ 
+2. $\ln y = 3x * ln(2x \sin x)$*
 # 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 $g(x) = x^2 + 3)$.
-7. First find the derivative of the outside function function ($f(x) = x^4$):
+3. First find the derivative of the outside function function ($f(x) = x^4$):
 $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
-8. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
+4. 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 $x^4$
 
@@ -187,7 +196,7 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$.
 
 Given these facts:
-9. Let $y$ be some function of $x$
-10. $\dfrac{d}{dx} x = 1$
-11. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
+5. Let $y$ be some function of $x$
+6. $\dfrac{d}{dx} x = 1$
+7. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\