Differentiate by Chain Rule
y=sin(x)+ln(x^2)+e^2x
Answers
Answer:
Step-by-step explanation:
We can find the derivative of this function implicitly. In other words, we will find the derivative of
y
, which will then allow us to find the derivative of
sin
(
x
)
ln
(
x
)
.
First, we want to get rid of the
ln
x
exponent. We can do that by taking the natural log of both sides and using a property of logarithms, that
ln
x
a
is equivalent to
a
ln
(
x
)
. Thus,
ln
y
=
ln
(
sin
x
)
ln
x
ln
y
=
ln
x
⋅
ln
(
sin
x
)
Now, we derive both sides. For the left side, we will have the derivative of
ln
y
=
1
y
, but we can't simply say that the derivative of
y
is 1 (using the chain rule). Rather, we say that it is
d
y
d
x
.
So, the left side of the equation now looks like this:
1
y
⋅
d
y
d
x
Now we take the derivative of the right side, which we can do using the product and chain rules. We get:
1
y
⋅
d
y
d
x
=
(
ln
(
sin
x
)
⋅
1
x
)
+
(
ln
x
⋅
1
sin
x
⋅
cos
x
)
1
y
⋅
d
y
d
x
=
ln
(
sin
x
)
x
+
ln
x
⋅
cos
x
sin
x
Remember that we're trying to solve for
d
y
d
x
. We can do this by multiplying both sides by
y
. Thus,
d
y
d
x
=
y
(
ln
(
sin
x
)
x
+
ln
x
⋅
cos
x
sin
x
)
Finally, to get our answer back into terms of
x
, we can replace the
y
on the right side of our derivative with
sin
x
ln
x
from the original function.
Our final answer is:
d
y
d
x
=
sin
(
x
)
ln
x
⋅
(
ln
(
sin
x
)
x
+
ln
x
⋅
cos
x
sin
x
)