Question

y=2sinx.logx find dy/dx​

Answer 1

Answer:

2(cosx.lnx+sinx/x)

Explanation:

hint : apply chain rule

Answer 2

Answer:

d y / d x = 2 [ log x . cos x + x⁻¹ . sin x ]

Explanation:

Given :

y = 2 sin x . log x

We have find d y / d x

We know :

Applying product rule :

d / d x [ ( f ( x ) / g ( x ) ] = g ( x ) . f' ( x ) + f ( x ) . g' ( x )

= > d / d x ( sin x ) = cos x

= > d / d x ( log x ) = 1 / x

Now diff. w.r.t. x :

= > d y / d x = 2 [ log x . ( sin x )' + sin x + ( log x )' ]

= > d y / d x = 2 [ log x . cos x + sin x . 1 / x ]

= > d y / d x = 2 [ log x . cos x + x⁻¹ . sin x ]

Hence we get required answer.

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