if u=x^4+y^4 then find ∂u/∂x
Answers
Step-by-step explanation:
Correct option is
A
2
The given information is: u=log(
x
2
+y
2
x
4
+y
4
)
Taking partial differentiation w.r.t. to x and y one at a time we get,
⇒
∂x
∂u
=
x
4
+y
4
x
2
+y
2
.
(x
2
+y
2
)
2
4x
3
(x
2
+y
2
)−2x(x
4
+y
4
)
⇒
∂x
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
4x
5
+4x
3
y
2
−2x
5
−2xy
4
⇒
∂y
∂u
=
x
4
+y
4
x
2
+y
2
.
(x
2
+y
2
)
2
4y
3
(x
2
+y
2
)−2y(x
4
+y
4
)
⇒
∂y
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
4y
5
+4x
2
y
3
−2y
5
−2x
4
y
Now we make calculation as follows,
⇒x
∂x
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
4x
6
+4x
4
y
2
−2x
6
−2x
2
y
4
⇒y
∂y
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
4y
6
+4x
2
y
4
−2y
6
−2x
4
y
2
⇒x
∂x
∂u
+y
∂y
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
4x
6
+4x
4
y
2
−2x
6
−2x
2
y
4
+
(x
4
+y
4
)(x
2
+y
2
)
4y
6
+4x
2
y
4
−2y
6
−2x
4
y
2
⇒x
∂x
∂u
+y
∂y
∂u
=
(x
4
+y
4
)(x
2
+y
2
)
2x
6
+2x
4
y
2
+2y
6
+2x
2
y
4
⇒x
∂x
∂u
+y
∂y
∂u
=
(x
2
+y
2
)(x
4
+y
4
)
2(x
2
+y
2
)(x
4
+y
4
)
⇒x
∂x
∂u
+y
∂y
∂u
=2