Question

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



Solve 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

Explanation:

What is the derivative of y=log [sin (logx)]?

Why do toppers choose BYJU'S for JEE preparation?

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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