WEBVTT
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for this problem where to find the derivative of Y
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equal to the integral firm cosine X two syntax of
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the natural log of one plus T V D V
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. Not to do this. You have to apply
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fundamental theorem of calculus which states that the derivative of
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the integral from A to let's say X of F
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F T D T. This is equal to F
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of X. Now in our given function the bands
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are both functions of X. So we need to
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partition this interval into two intervals in which one of
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the endpoint is a constant. That's e knows that
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in the interval co sign X to cynics there is
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some value A in between it. So we can
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say that this is equal to um the interval from
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co sign X to A. And then union from
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A to sign of X. And so we can
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rewrite the function into why that's equal to the integral
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from cosign X to E. Of L N.
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Of one plus two V. D V plus we
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have the integral from a two Synnex of L N
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. Of one plus T V D. V.
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And because you're following fundamental theorem of calculus, we
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need to we write the first integral so that the
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function the boundary function will go up and then the
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boundary they will be the bottom boundary. And so
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we rewrite this into why is just equal to the
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negative of the integral from a to co sign X
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. Of the natural log of one plus two Vdv
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plus we have the integral from A to cynics,
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ah the natural log of one plus two V.
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D. V. And so by fundamental theorem of
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calculus and chain rule we have why prime? This
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is equal to the negative of the natural log of
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one plus two times cosine x times the derivative of
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cosine X. And then plus we have um Ln
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of one plus two times cynics times the derivative of
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Sine X. And simplifying this, we get negative
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Ln of one plus to cosign X times the derivative
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Goldstein exit just negative Synnex plus we have Ln of
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one plus two syntax times the derivative of Sine X
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. Which is co sign X. And so we
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get sin X times Ln of one plus to co
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sign X plus. Co sign X times Ln of
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one plus to sign Mhm. And so this is
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the derivative of the function.