differentiate w.r.t.x x^x + a^a
Answers
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Step-by-step explanation:
Let y=x
x
+x
a
+a
x
+a
a
Also let, x
x
=u,x
a
=v,a
x
=w, and a
a
=s
∴y=u+v+w+s
⇒
dx
dy
=
dx
du
+
dx
dv
+
dx
dw
+
dx
ds
.....(1)
u=x
x
⇒logu=logx
x
⇒logu=xlogx
Differentiating both sides with respect to x, we obtain
u
1
dx
du
=logx.
dx
d
(x)+x.
dx
d
(logx)
⇒
dx
du
=u[logx.1+x
x
1
]
⇒
dx
du
=x
x
[logx+1]=x
x
(1+logx) .....(2)
v=x
a
∴
dx
dv
=
dx
d
(x
a
)
⇒
dx
dv
=ax
a−1
.....(3)
w=a
x
⇒logw=loga
x
⇒logw=xloga
Differentiating both sides with respect to x, we obtain
w
1
.
dx
dw
=loga.
dx
d
(x)
⇒
dx
dw
=wloga
⇒
dx
dw
=a
x
loga .....(4)
s=a
a
Since a is constant, a
a
is also a constant.
∴
dx
ds
=0 .....(5)
From (1), (2), (3), (4) and (5) we obtain
dx
dy
=x
x
(1+logx)+ax
a−1
+a
x
loga+0
=x
x
(1+logx)+ax
a−1
+a
x
loga
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