Differentiate x x cos w. r. t. x (cos x) .
Answers
Answer:
Solution To Question ID 421211
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Answer
Let y=x
xcosx
+
x
2
−1
x
2
+1
Also, let u=x
xcosx
and v=
x
2
−1
x
2
+1
∴y=u+v
⇒
dx
dy
=
dx
du
+
dx
dv
...(1)
u=x
xcosx
⇒logu=log(x
xcosx
)
⇒logu=xcosxlogx
Differentiating both sides with respect to x, we obtain
u
1
dx
du
=
dx
d
(x).cosx.logx+x
dx
d
(cosx).logx+xcosx.
dx
d
(logx)
⇒
dx
du
=u[1.cosx.logx+x.(−sinx)logx+xcosx.
x
1
]
⇒
dx
du
=x
xcosx
(cosx.logx−xsinxlogx+cosx)
⇒
dx
du
=x
xcosx
[cosx(1+logx)−xsinxlogx] ...(2)
v=
x
2
−1
x
2
+1
⇒logv=log(x
2
+1)−log(x
2
−1)
Differentiating both sides with respect to x, we obtain
v
1
dx
dv
=
x
2
+1
2x
−
x
2
−1
2x
⇒
dx
dv
=v[
(x
2
+1)(x
2
−1)
2x(x
2
−1)−2x(x
2
+1)
]
⇒
dx
dv
=
x
2
−1
x
2
+1
×[
(x
2
+1)(x
2
−1)
−4x
]
⇒
dx
dv
=
(x
2
−1)
2
−4x
...(3)
From (1), (2) and (3), we obtain
⇒
dx
dy
=x
xcosx
[cosx(1+logx)−xsinxlogx]−
(x
2
−1)
2
4x