Question

if y= (sin x)^sin x then find dy/dx​

Answer 1

Answer:

y = (sin x)^sin x

applying log on both sides,

log y = log (sin x)^sin x [ product rule ]

=> log y = sin x log sin x [ log m^n = n log m ]

differentiating both sides wet x,

1/y log y = sin x d/dx(log sin x) + log sin x d/dx(sinx)

=> 1/y dy/dx = sin x * 1/sin x* cosx + log sin x*cos x [chain rule log sin x=1/sin x*cosx ]

=> 1/y dy/dx = cos x + cos x log sin x

=> dy/dy = y [ cos x + cos x log sin x ]

=> dy/dx=(sin x)^sin x [cos x+cos x log sin x ]

Answer 2

Answer:

( sin x )^sin x ( cos x . ㏑ sin x + cos x )

Step-by-step explanation:

Given :

y = ( sin x )^sin x

Taking ㏑ both side we get :

= > ㏑ y = ㏑ ( sin x )^sin x

= > ㏑ y = ( sin x ) ㏑ sin x

Diff. w.r.t. x :

= > 1 / y ( d y / d x ) = ㏑ sin x . ( sin x )' + sin x ( ㏑ sin x )'

= >  1 / y ( d y / d x ) =  cos x . ㏑ sin x + 1 / sin . ( sin x ) . cos x

= >  1 / y ( d y / d x ) =  cos x . ㏑ sin x + cos x

Putting value of y = ( sin x )^sin x

= > d y / d x =  ( sin x )^sin x ( cos x . ㏑ sin x + cos x )

Hence we get required answer!