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Let's get some more practice
doing implicit differentiation.
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So let's find the derivative
of y with respect to x.
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We're going to assume
that y is a function of x.
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So let's apply our
derivative operator
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to both sides of this equation.
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So let's apply our
derivative operator.
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And so first, on
the left hand side,
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we essentially are just going
to apply the chain rule.
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First we have the
derivative with respect
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to x of x minus y squared.
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So the chain rule
tells us this is
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going to be the derivative
of the something squared
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with respect to the
something, which is just
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going to be 2 times x
minus y to the first power.
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I won't write the
1 right over there.
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Times the derivative of the
something with respect to x.
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Well, the derivative of x
with respect to x is just 1,
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and the derivative
of y with respect
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to x, that's what
we're trying to solve.
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So it's going to
be 1 minus dy dx.
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Let me make it a
little bit clearer
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what I just did right over here.
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This right over here
is the derivative
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of x minus y squared with
respect to x minus y.
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And then this right over
here is the derivative
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of x minus y with respect to x.
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Just the chain rule.
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Now let's go to the right
hand side of this equation.
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This is going to be equal to
the derivative of x with respect
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to x is 1.
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The derivative of y
with respect to x.
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We're just going to write
that as the derivative of y
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with respect to x.
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And then finally, the
derivative with respect
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to x of a constant, that's
just going to be equal to 0.
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Now let's see if we can
solve for the derivative of y
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with respect to x.
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So the most obvious thing to do.
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Let's make it clear.
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This right over here, I
can rewrite as 2x minus 2y.
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So let me do that so
I can save some space.
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This is 2x minus 2y If
I just distribute the 2.
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And now I can distribute
the 2x minus 2y
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onto each of these terms.
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So 2x minus 2y times 1 is
just going to be 2x minus 2y.
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And then 2x minus 2y
times negative dy dx,
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that's just going to be
negative 2x minus 2y.
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Or we could write that as
2y minus 2x times dy dx.
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Is equal to 1 plus dy dx.
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I'll do all my dy
dx's in orange now.
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1 plus dy dx.
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So now there's a
couple of things
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that we could attempt to do.
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We could subtract 2x
minus 2y from both sides.
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So let's do that.
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So let's subtract 2x
minus 2y from both sides.
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So over here, we're going
to subtract 2x minus
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2y from that side.
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And then we could also subtract
a dy dx from both sides,
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so that all of our dy dx's
are on the left hand side,
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and all of our non dy dx's
are on the right hand side.
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So let's do that.
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So we're going to subtract a
dy dx on the right and a dy dx
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here on the left.
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And so what are we left with?
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Well, on the left hand
side, these cancel out.
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And we're left with 2y minus
2x dy dx minus 1 dy dx,
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or just minus a dy dx.
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Let me make it clear.
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We could write this
as a minus 1 dy dx.
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So this is we can
essentially just add
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these two coefficients.
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So this simplifies to 2y minus
2x minus 1 times the derivative
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of y with respect
to x, which is going
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to be equal to-- on this
side, this cancels out.
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We are left with 1
minus 2x plus 2y.
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So let me write it that way.
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Or we could write
this as-- so negative,
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negative 2y is
just a positive 2y.
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And then we have minus 2x.
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And then we add that 1, plus 1.
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And now to solve
for dy dx, we just
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have to divide both sides
by 2y minus 2x minus 1.
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And we are left
with-- we deserve
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a little bit of a drum
roll at this point.
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As you can see, the
hardest part was really
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the algebra to solve for dy dx.
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We get the derivative
of y with respect
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to x is equal to
2y minus 2x plus 1
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over 2y minus 2x minus 1.