x^y+y^x=a, find the value of dy/dx
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Answer:
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Step-by-step explanation:
Solution
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We have,
y
x
+x
y
+x
x
=a
b
⇒ e
logy
x
+e
logx
y
+e
logx
x
=a
b
⇒ e
xlogy
+e
ylogx
+e
xlogx
=a
b
Differentiating both sides with respect to x, we get
dx
d
(e
xlogy
)+
dx
d
(e
ylogx
)+
dx
d
(e
xlogx
)=
dx
d
a
b
)
⇒ e
xlogy
dx
d
(xlogy)+e
ylogx
dx
d
(ylogx)+e
xlogx
dx
d
(xlogx)=0
⇒ y
x
(logy+
y
x
dx
dy
)+x
y
(
dx
dy
logx+
x
y
)+x
x
(1+logx)=0
⇒
dx
dy
(xy
x−1
+x
y
logx)=−{y
x
logy+y x
y−1
+x
x
(1+logx)}
⇒
dx
dy
=−
xy
x−1
+x
y
logx
{y
x
logy+y x
y−1
+x
x
(1+logx)}
Answered by
0
Answer:
If y^x + x^y + x^x = a^b , find dydx
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