If y = log(sin (log x)] then find
Dy/dx
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So given, y = log[sin(logx)]
Let's first take z = sin(logx)
So we have y = logz
Use chain rule now
dy/dx = dy/dz * dz/dx = dlogz / dz * dz/dx
= 1/z * dz/dx
Now again use chain rule for dz/dx, take t = logx
dy/dx = 1/sin(logx) * dsint / dt * dt/dx
dy/dx = 1/sin(logx) * cos(logx) * dlogx / dx
dy/dx = cotx / x
Hope it helped! :)
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