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Step-by-step explanation:
Let y=x
x
−2
sinx
Also, let x
x
=u and 2
sinx
=v
∴y=u−v
⇒
dx
dy
=
dx
du
−
dx
dv
u=x
x
Taking logarithm on both the sides, we obtain
logu=xlogx
Differentiating both sides with respect to x, we obtain
u
1
dx
du
=[
dx
d
(x)×logx+x×
dx
d
(logx)]
⇒
dx
du
=u[1×logx+x×
x
1
]
⇒
dx
du
=x
x
(1+logx)
v=2
sinx
Taking logarithm on both the sides with respect to x, we obtain
logv=sinxlog2
Differentiating both sides with respect to x, we obtain
v
1
.
dx
dv
=log2.
dx
d
(sinx)
⇒
dx
dv
=vlog2cosx
⇒
dx
dv
=2
sinx
cosxlog2
∴
dx
dy
=x
x
(1+logx)−2
sinx
cosxlog2
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