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Given that π₯ equals five π‘ minus four times the natural log of π‘ and π¦ equals four π‘ plus five times the natural log of three π‘, find dπ¦ by dπ₯.
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Here weβve been given a pair of parametric equations.
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These are equations for π₯ and π¦ in terms of a third parameter.
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Thatβs π‘.
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And weβre looking to find the derivative of π¦ with respect to π₯.
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And so, we recall that as long as dπ₯ by dπ‘ is not equal to zero, the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
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And so, it should be quite clear to us that weβre going to need to differentiate each of our respective equations with respect to π‘.
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And weβll do this term by term.
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The first term in our equation for π₯ is five π‘.
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Now, the derivative of five π‘ with respect to π‘ is five.
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Next, we recall that the derivative of the natural log of ππ₯, where π is some constant, with respect to π₯ is one over π₯.
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So, the derivative of the natural log of π‘ is one over π‘.
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And so, the derivative of negative four times the natural log of π‘ is negative four over π‘.
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Letβs repeat this process for dπ¦ by dπ‘.
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The derivative of four π‘ with respect to π‘ is four.
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Then, the derivative of the natural log of three π‘ must be one over π‘.
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So, five times the natural log of three π‘ is five over π‘. dπ¦ by dπ₯ is the quotient of these.
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Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
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So, thatβs four plus five over π‘ divided by five minus four over π‘.
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Now, at the moment, this doesnβt look particularly nice, so weβre going to deal with the fractions in our numerator and denominator.
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Weβre going to multiply through by π‘.
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Then, our numerator becomes four π‘ plus five, and our denominator becomes five π‘ minus four.
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And so, weβre done.
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We found dπ¦ by dπ₯.
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Itβs four π‘ plus five over five π‘ minus four.