Question Number 103149 by bemath last updated on 13/Jul/20

using first principal   y = ln (sin (√x)) →y′ = ?

Answered by bobhans last updated on 13/Jul/20

y′ = lim_(△x→0) ((ln(sin (√(x+△x)))−ln(sin (√x)))/(△x))  y′=lim_(△x→0) ((ln(((sin (√(x+△x)))/(sin (√x)))))/(△x))

Answered by mathmax by abdo last updated on 14/Jul/20

y(x) =ln(sin((√x))) ⇒y^′ (x) =(((sin((√x)))^′ )/(sin((√x)))) =((cos((√x)))/(2(√x)sin((√x)))) =(1/(2(√x)))cotan((√x))