WEBVTT
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here we have a composite function, a function inside
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of function, and so we're going to find the
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derivative using the chain rule. But on the inside
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we have a quotient. So we're going to end
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up using the quotient rule at some point as well
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. So, first of all, for the derivative
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of the outside, the derivative of co sign is
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negative sign. So have negative sign of the quotient
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that we see. And then we're gonna multiply that
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by the derivative of the quotient. So we have
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the bottom one plus e to the two x times
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the derivative of the top and the derivative of one
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minus C to the two X would be minus or
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negative each of the two x Times two. We
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had to use the chain rule on that as well
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. The derivative of two X is too. So
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that's the bottom times the derivative of the top minus
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the top one minus each of the two x times
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, the derivative of the bottom. And that would
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be each of the two x Times two. And
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that's gonna be over the bottom squared. So over
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one plus e to the two x squared now what
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we want to do is see what we can do
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to simplify that second numerator, and I suspect that
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what we might do is factor some things out.
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So let's see what we could factor out of both
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. So we have a negative that we could factor
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out of both. We have a two that we
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could factor out of both, and we have a
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knee to the two X that we could factor out
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of both. So when we do that, we're
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going to have because we're factoring out of negative.
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It's going to cancel with the negative on the outsides
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. We're gonna have positive to each of the two
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X times the sign of one minus each of the
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two X over one plus each of the two X
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times. Now let's see what we have left.
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In that quotient, we have one plus E to
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the two x plus one minus each of the two
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X over one plus each of the two x squared
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. Now look what happens to the numerator of that
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quotient. The each of the two x and the
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minus each of the two x cancel, leaving us
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with just one plus one, and we know that
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that is to so let's rewrite her answer did ourselves
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a little more space. So what you see here
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is what we just had a moment ago, and
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we're going to take this to and multiply it by
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two. The two that we have in the front
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. So we have four e to the two x
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times. Or we could just write that over the
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denominator that we have here over one plus e to
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the two x squared times. A sign of the
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quotient. One minus each of the two x over
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one plus each of the two x.