Question

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

Please please

Answer 1

Explanation:

Explanation:

Answer:

( cos ( ㏒ x ) ) / ( x . sin ( ㏒ x ) )

Step-by-step explanation:

Let :

y = ㏒ ( sin ( ㏒ x ) )   ,   x > 0

# ㏒ x = ㏒_e x = ㏑ x

We know :

= > ( ㏑ x )' = 1 / x

= > y' = 1 / sin ( ㏒ x ) . ( sin ( ㏒ x ) )'

= >  y' = 1 / sin ( ㏒ x ) . ( cos ( ㏒ x ) ) . ( ㏒ x )'

= > y' = 1 / sin ( ㏒ x ) . ( cos ( ㏒ x ) ) . 1 / x

= > y'  = ( cos ( ㏒ x ) ) / ( x . sin ( ㏒ x ) )

Hence we get required answer!

Answer 2

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

By chain rule ,

$\sf \mansto \frac{dy}{dx} = \frac{1}{Sin \{Log(x) \}} \times \frac{d \{Sin \{Log(x) \} \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} =\frac{1}{Sin \{Log(x) \}} \times Cos \{Log(x) \} \times \frac{d \{Log(x) \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{Sin \{Log(x) \}} \times Cos \{Log(x) \} \times \frac{1}{x} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{Cos \{Log(x) \}}{x.Sin \{Log(x) \}}</p><p>$

Remmember :

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