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\( \cot v = \csc ^ { 2 } v \frac { d } { d x } v \)

Ask by Rogers Harrington. in Mexico
Mar 06,2025

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Answer

La derivada de \( v \) con respecto a \( x \) es \( \frac{dv}{dx} = \sin v \cos v \) o \( \frac{dv}{dx} = \frac{1}{2}\sin 2v \).

Solution

A continuación, resolveremos paso a paso la ecuación \[ \cot v = \csc^2 v \frac{d}{dx} v \] **Paso 1. Escribir las funciones en términos de seno y coseno** Recordamos que \[ \cot v = \frac{\cos v}{\sin v} \quad \text{y} \quad \csc v = \frac{1}{\sin v}. \] Entonces, \[ \csc^2 v = \frac{1}{\sin^2 v}. \] **Paso 2. Sustituir en la ecuación original** Reemplazando en la ecuación dada, tenemos \[ \frac{\cos v}{\sin v} = \frac{1}{\sin^2 v}\frac{d}{dx}v. \] **Paso 3. Despejar \(\frac{d}{dx}v\)** Multiplicamos ambos lados de la ecuación por \(\sin^2 v\): \[ \frac{\cos v}{\sin v} \sin^2 v = \frac{d}{dx}v. \] Notamos que \[ \frac{\cos v}{\sin v} \sin^2 v = \cos v \sin v. \] Así, obtenemos \[ \frac{d}{dx}v = \cos v \sin v. \] **Paso 4. Interpretación del resultado** La derivada de \( v \) con respecto a \( x \) es \[ \frac{dv}{dx} = \sin v \cos v. \] Opcionalmente, usando la identidad del ángulo doble \[ \sin 2v = 2 \sin v \cos v, \] podemos escribir la derivada en forma alternativa: \[ \frac{dv}{dx} = \frac{1}{2}\sin 2v. \] **Conclusión** Hemos demostrado que, a partir de la ecuación \[ \cot v = \csc^2 v \frac{d}{dx}v, \] la derivada de \( v \) respecto a \( x \) es \[ \frac{dv}{dx} = \sin v \cos v \quad \text{o} \quad \frac{dv}{dx} = \frac{1}{2}\sin 2v. \]

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Bonus Knowledge

Did you know that the cotangent function, \( \cot v \), is actually the reciprocal of the tangent function? This means that while \( \tan v = \frac{\sin v}{\cos v} \), then \( \cot v = \frac{\cos v}{\sin v} \). It's like a see-saw where one side goes up as the other goes down! This identity you’re exploring taps into the intricate relationships between trigonometric functions. Also, this equation resembles the structure of implicit differentiation used in calculus, where we derive functions regarding dependent and independent variables. When you manipulate trigonometric functions or implicit relationships, it's crucial to keep track of function derivatives. A common mistake is not paying attention to the chain rule during such derivations. Keep practicing, and you’ll master these!

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