if y=(sinx ) ^ x find dy/dx
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Step-by-step explanation:
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Answer:
( sin x )^x ( x . cot x + ㏑ sin x )
Step-by-step explanation:
Given :
y = ( sin x )^x
Taking ㏑ both side we get :
= > ㏑ y = ㏑ ( sin x )^x
= > ㏑ y = x . ㏑ sin x
We know :
( ㏑ x )' = 1 / x
( sin x )' = cos x
Diff. w.r.t. x :
= > 1 / y ( d y / d x ) = x ( ㏑ sin x )' + ㏑ sin x ( x )'
= > 1 / y ( d y / d x ) = x ( ㏑ sin x )' + ㏑ sin x
= > 1 / y ( d y / d x ) = x / sin x . ( sin x )' + ㏑ sin x
= > 1 / y ( d y / d x ) = x / sin x . ( cos x ) + ㏑ sin x
= > 1 / y ( d y / d x ) = x . cos x / sin x + ㏑ sin x
= > 1 / y ( d y / d x ) = x . cot x + ㏑ sin x
Putting value of y = ( sin x )^x
= > d y / d x = ( sin x )^x ( x . cot x + ㏑ sin x )
Hence we get required answer!
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