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Exercise

$\frac{\cos\left(2x-1\right)}{\sin\left(2x\right)}$

Step-by-step Solution

1

Utilizando la identidad del coseno de la suma de dos ángulos: $\cos(\alpha\pm\beta)=\cos(\alpha)\cos(\beta)\mp\sin(\alpha)\sin(\beta)$, donde el ángulo $\alpha$ equivale a $2x$, y el ángulo $\beta$ equivale a $-1$

$\frac{\cos\left(-1\right)\cos\left(2x\right)-\sin\left(-1\right)\sin\left(2x\right)}{\sin\left(2x\right)}$

Final answer to the exercise

$\frac{\cos\left(-1\right)\cos\left(2x\right)-\sin\left(-1\right)\sin\left(2x\right)}{\sin\left(2x\right)}$

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Similar Questions Solved

$derivdef\left(x<y<0\right)$

Main Topic: Definition of Derivative

Resolution of derivatives using the definition of the derivative, which is the limit of difference quotients of real numbers.

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