Question

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

Answer 1

Answer:

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

Explanation:

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Answer 2

Let , y = Log{Sin(Log(x)}

By chain rule ,

$\begin{gathered}\sf \mapsto \frac{dy}{dx} = \frac{1}{Sin \{Log(x) \}} \times \frac{d \{Sin \{Log(x) \} \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} =\frac{1}{Sin \{Log(x) \}} \times Cos \{Log(x) \} \times \frac{d \{Log(x) \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{Sin \{Log(x) \}} \times Cos \{Log(x) \} \times \frac{1}{x} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{Cos \{Log(x) \}}{x.Sin \{Log(x) \}}\end{gathered}$

Remmember :

$\begin{gathered}\star \sf \: \: \frac{d \{Sin(x) \}}{x} = Cos(x) \\ \\ \star \: \: \sf \frac{d \{Log(x) \}}{dx} = \frac{1}{x}\end{gathered}$