Question

find the differential co-effecient of y=. x^x logx (sinx)^sinx
help please​

Answer 1

Step-by-step explanation:

ANSWER

We are given that y=(sinx)

logx

+x

sinx

We have to find

dx

dy

,

let y

1

=(sinx)logx & y

2

=x

sinx

dx

dy

1

+

dx

dy

2

=

dx

dy

now, we have y

1

=(sinx)logx

logy

1

=(logx)(log(sinx))

y

1

1

dx

dy

1

=

x

1

(log(sinx))+

(sinx)

logx

(cosx)

dx

dy

1

=y

1

[

x

log(sinx)

+(logx)cotx]

now, y

2

=x

sinx

⇒logy

2

=(sinx)(logx)

y

2

1

dx

dy

2

=

x

sinx

+(cosx)(logx)

dx

dy

2

=y

2

[

x

sinx

+(logx)(cosx)]

dx

dy

=

dx

dy

1

+

dx

dy

2

dx

dy

=(sinx)

logx

[

x

log(sinx)

+(logx)(cotx)]+x

sinx

[

x

sinx

+(logx)(cosx)]

Answer 2

Answer:

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