solve logarithmic differentiation, find dy/dx if y=(cosh e^x + coth^4 x)^(log3 (x))
Answers
Answer:
Solution
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Correct option is
A
cosh
2
xsinh
2
x
−(sinh
3
x+cosh
3
x)
We have,
dx
d
⎩
⎪
⎨
⎪
⎧
e
2
1
log(1−tanh
2
x)
+3
2
1
log
3
(coth
2
x−1)
⎭
⎪
⎬
⎪
⎫
=
dx
d
{e
log(
1−tanh
2
x
)
+3
log
3
(
coth
2
x−1
)
}
=
dx
d
{
1−tanh
2
x
+
coth
2
x−1
}
={
2
1−tanh
2
x
1
×(0−2tanhx(1−tanh
2
x))+
2
coth
2
x−1
1
×(2cothx(1−coth
2
x)−0)}
={
1−tanh
2
x
−tanhx(1−tanh
2
x)
−
coth
2
x−1
cothx(coth
2
x−1)
}
={−tanhx(
1−tanh
2
x
)−cothx(
coth
2
x−1
)}
=
⎩
⎪
⎨
⎪
⎧
−
coshx
sinhx
⎝
⎛
1−
cosh
2
x
sinh
2
x
⎠
⎞
−
sinhx
coshx
⎝
⎛
sinh
2
x
cosh
2
x
−1
⎠
⎞
⎭
⎪
⎬
⎪
⎫
=−{
cosh
2
x
sinhx
(
cosh
2
x−sinh
2
x
)+
sinh
2
x
coshx
(
cosh
2
x−sinh
2
x
)}
=−{
cosh
2
x
sinhx
+
sinh
2
x
coshx
}(
cosh
2
x−sinh
2
x
)
=−(
sinh
2
xcosh
2
x
sinh
3
x+cosh
3
x
)(
cosh
2
x−sinh
2
x
)
We know that
cosh
2
x−sinh
2
x=1
Therefore,
=−(
sinh
2
xcosh
2
x
sinh
3
x+cosh
3
x
)×1
=−(
sinh
2
xcosh
2
x
sinh
3
x+cosh
3
x
)
Hence, this is the answer