differentiate the following wrt x 1/x^3/2
Answers
Answer:
Let y=x
x
2
−3
+(x−3)
x
2
Also, let u=x
x
2
−3
and v=(x−3)
x
2
∴y=u+v
Differentiating both sides with respect to x, we obtain
dx
dy
=
dx
du
+
dx
dv
.....(1)
u=x
x
2
−3
∴logu=log(x
x
2
−3
)
⇒logu=(x
2
−3)logx
Differentiating with respect to x, we obtain
u
1
dx
du
=logx.
dx
du
=logx.
dx
d
(x
2
−3)+(x
2
−3).
dx
d
(logx)
⇒
u
1
dx
du
=logx.2x+(x
2
−3).
x
1
dx
du
=x
x
2
−3
[
x
x
2
−3
+2xlogx]
Also,
v=(x−3)
x
2
∴logv=log(x−3)
x
2
⇒logv=x
2
log(x−3)
Differentiating both sides with respect to x, we obtain
v
1
dx
dv
=log(x−3).
dx
d
(x
2
)+x
2
.
dx
d
[log(x−3)]
⇒
v
1
dx
dv
=log(x−3).2x+x
2
.
x−3
1
.
dx
d
(x−3)
⇒
dx
dv
=v[2xlog(x−3)+
x−3
x
2
.1]
⇒
dx
dv
=(x−3)
x
2
[
x−3
x
2
+2xlog(x−3)]
Substituting the expressions of
dx
du
and
dx
dv
in equation (1), we obtain
dx
dy
=x
x
2
−3
[
x
x
2
−3
+2xlogx]+(x−3)
x
2
[
x−3
x
2
+2xlog(x−3)]
Step-by-step explanation: