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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. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\