Solución detallada
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No logro encontrar los pasos en la búsqueda de esta derivada.
Perola derivada
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Simplificamos:
Respuesta:
x - 1 /2*(x - 1)*cos(2*x) \
sin (2*x)*|------------------ + log(sin(2*x))|
\ sin(2*x) /
$$\left(\frac{2 \left(x - 1\right) \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right) \sin^{x - 1}{\left(2 x \right)}$$
/ 2 2 \
-1 + x | /2*(-1 + x)*cos(2*x) \ 4*cos(2*x) 4*cos (2*x)*(-1 + x)|
sin (2*x)*|4 + |------------------- + log(sin(2*x))| - 4*x + ---------- - --------------------|
| \ sin(2*x) / sin(2*x) 2 |
\ sin (2*x) /
$$\left(- 4 x - \frac{4 \left(x - 1\right) \cos^{2}{\left(2 x \right)}}{\sin^{2}{\left(2 x \right)}} + \left(\frac{2 \left(x - 1\right) \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right)^{2} + 4 + \frac{4 \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\right) \sin^{x - 1}{\left(2 x \right)}$$
/ 3 2 / 2 \ 3 \
-1 + x | /2*(-1 + x)*cos(2*x) \ 12*cos (2*x) /2*(-1 + x)*cos(2*x) \ | cos(2*x) cos (2*x)*(-1 + x)| 16*cos (2*x)*(-1 + x) 16*(-1 + x)*cos(2*x)|
sin (2*x)*|-12 + |------------------- + log(sin(2*x))| - ------------ - 12*|------------------- + log(sin(2*x))|*|-1 + x - -------- + ------------------| + --------------------- + --------------------|
| \ sin(2*x) / 2 \ sin(2*x) / | sin(2*x) 2 | 3 sin(2*x) |
\ sin (2*x) \ sin (2*x) / sin (2*x) /
$$\left(\frac{16 \left(x - 1\right) \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \frac{16 \left(x - 1\right) \cos^{3}{\left(2 x \right)}}{\sin^{3}{\left(2 x \right)}} + \left(\frac{2 \left(x - 1\right) \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right)^{3} - 12 \left(\frac{2 \left(x - 1\right) \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right) \left(x + \frac{\left(x - 1\right) \cos^{2}{\left(2 x \right)}}{\sin^{2}{\left(2 x \right)}} - 1 - \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\right) - 12 - \frac{12 \cos^{2}{\left(2 x \right)}}{\sin^{2}{\left(2 x \right)}}\right) \sin^{x - 1}{\left(2 x \right)}$$