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Suppose π₯ equals nine π minus five and π¦ equals four root π plus six π squared.
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Find dπ¦ by dπ₯ when π is equal to four.
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Here we have a pair of parametric equations expressed in terms of a third parameter, π.
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We recall that, for two differentiable functions, π and π, where π₯ is equal to π of π‘ and π¦ is equal to π of π‘, dπ¦ by dπ₯ is found by dividing dπ¦ by dπ‘ by dπ₯ by dπ‘.
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Now of course, our functions are in terms of π.
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So here we can say that dπ¦ by dπ₯ must be equal to dπ¦ by dπ over dπ₯ by dπ.
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And then itβs quite clear weβre going to need to differentiate both of our functions with respect to π.
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Well, the derivative of nine π is nine.
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And the derivative of negative five is zero.
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So dπ₯ by dπ is nine.
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Letβs write four root π as four times π to the power of one-half.
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Then the derivative of four π to the power of one-half is a half times four π to the power of negative one-half, which simplifies to two π to the power of negative one-half.
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Then the derivative of six π squared is two times six π, which is 12π. dπ¦ by dπ₯ is then the quotient of these.
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Itβs two π to the power of negative one-half plus 12π all over nine.
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Now of course, weβre looking to find the value of the derivative when π is equal to four.
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So letβs substitute π equals four into our function.
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Thatβs two times four to the power of negative one-half plus 12 times four over nine.
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Well, four to the power of negative one-half is one over the square root of four, or one-half.
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And two times one-half is just [one].
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We know that 12 times four is 48 and one plus 48 is 49.
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And so when π is equal to four, dπ¦ by dπ₯ is equal to 49 over nine.