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Find the slope of the tangent to the curve five π₯ over two π¦ minus two π¦ over π₯ equals negative four at the point two, five.
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We should recall that to find the slope of the tangent to a curve, we need to evaluate the derivative of that function at a given point.
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Now, weβre most used to evaluating derivatives of explicitly defined functions of π₯.
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In this case though, weβre looking at a function that is defined implicitly.
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And so, weβre going to use implicit differentiation to evaluate the derivative.
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This is just a special case of the chain rule.
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And it says that the derivative of some function in π¦ with respect to π₯ is equal to the derivative of that function with respect to π¦ times dπ¦ by dπ₯.
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And so, letβs look at the equation of our curve.
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Itβs five π₯ over two π¦ minus two π¦ over π₯ equals negative four.
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And weβll just begin by differentiating each side of this equation with respect to π₯.
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We recall that the derivative of the sum or difference of two or more functions is equal to the sum or difference of their respective derivatives.
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And so, weβre going to differentiate five π₯ over two π¦, two π¦ over π₯, and negative four.
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Well, in fact, the derivative of a constant is zero.
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So, the right-hand side here becomes zero.
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But how do we differentiate five π₯ over two π¦ and two π¦ over π₯?
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Well, we use the quotient rule.
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This says that the derivative of the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.
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Letβs begin by looking at the derivative of five π₯ over two π¦.
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Weβll let π’ be equal to five π₯, thatβs the numerator, and π£ be equal to two π¦, thatβs the denominator.
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Then, dπ’ by dπ₯, the first derivative of five π₯, is simply five.
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But what about dπ£ by dπ₯?
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Well, according to the earlier formula, which was just a special case of the chain rule, we multiply the derivative of the function with respect to π¦, so thatβs two, by dπ¦ by dπ₯.
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And so, dπ£ by dπ₯ is simply two dπ¦ by dπ₯.
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Then, the derivative of five π₯ over two π¦ is π£ times dπ’ by dπ₯, so thatβs two π¦ times five, minus π’ times dπ£ by dπ₯, so thatβs five π₯ times two dπ¦ by dπ₯, all over π£ squared, so two π¦ squared.
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If we simplify a little, we get 10π¦ minus 10π₯ dπ¦ by dπ₯ all over four π¦ squared.
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And then, we should see that we can divide through by a common factor of two.
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And we found the derivative of the first part of our expression on the left-hand side.
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Itβs five π¦ minus five π₯ dπ¦ by dπ₯ all over π¦ squared.
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Weβre now going to repeat this process for the derivative of two π¦ over π₯.
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This time, we let π’ be equal to two π¦ and π£ be equal to π₯.
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Of course, we know that the derivative of π₯ with respect to π₯ is just one.
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We already evaluated the derivative of two π¦.
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But as a reminder, itβs the derivative of two π¦ with respect to π¦ which is two times dπ¦ by dπ₯.
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Letβs substitute everything we have into our formula for the quotient rule.
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And we get two π₯ dπ¦ by dπ₯ minus two π¦ over π₯ squared.
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And so, we find that when we differentiate both sides of our equation with respect to π₯, we get five π¦ minus five π₯ dπ¦ by dπ₯ over π¦ squared minus two π₯ dπ¦ by dπ₯ minus two π¦ over π₯ squared equals zero.
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But what on earth are we going to do with this?
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Remember, weβre trying to find the slope of the tangent, so we need to evaluate dπ¦ by dπ₯ at our point two, five.
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The problem is, we havenβt currently got an expression for dπ¦ by dπ₯.
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So, letβs see if we can find a way to simplify a little.
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With our two fractions, weβre going to divide each bit of the numerator by the denominator.
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So, we divide five π¦ by π¦ squared to get five over π¦.
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And we divide negative five π₯ dπ¦ by dπ₯ by π¦ squared to get negative five π₯ over π¦ squared dπ¦ by dπ₯.
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Similarly, we divide negative two π₯ dπ¦ by dπ₯ by π₯ squared to get negative two over π₯ dπ¦ by dπ₯ and negative negative two π¦ by π₯ squared to get positive two π¦ over π₯ squared.
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We now see that we can factor dπ¦ by dπ₯.
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And with those terms, we get dπ¦ by dπ₯ times negative five π₯ over π¦ squared minus two over π₯.
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At the same time, letβs subtract five over π¦ and two π¦ over π₯ squared from both sides.
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And so, we have this given expression.
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We now should be able to see that we can divide through by negative five π₯ over π¦ squared minus two over π₯.
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We might also notice that every single term has a factor of negative one.
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So, weβre going to divide through by negative one at the same time.
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And so, we have a rather nasty-looking expression for the first derivative of π¦ with respect to π₯.
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Itβs five over π¦ plus two π¦ over π₯ squared all over five π₯ over π¦ squared plus two over π₯.
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Normally, we might look to simplify this a little, but remember weβre trying to find the slope of the tangent at a given point.
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In fact, itβs the point where π₯ equals two and π¦ equals five.
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So, letβs substitute π₯ equals two and π¦ equals five into our expression for the derivative.
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The numerator becomes five over five plus two times five over two squared, which simplifies to one plus five over two.
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Then, our denominator is five times two over five squared plus two over two, which is two-fifths plus one.
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One plus five over two is seven over two.
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And two-fifths plus one is seven-fifths.
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So, we have seven over two divided by seven-fifths.
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And then, we might recall that to divide by a fraction, we multiply the reciprocal of that same fraction.
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So, itβs seven over two times five-sevenths.
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And then the sevens cancel, to get five over two.
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And so, weβre done.
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We found the slope of the tangent to our curve at the point two, five.
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Itβs five over two, or five-halves.