Question

The derivative of log (sin(logx) (x > 0) ​

Khhh

Answer 1

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! :)

Answer 2

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! :)