Video: Finding the Equation of the Tangent to a Curve Defined Implicitly at a Given Point Using Implicit Differentiation

Find the slope of the tangent to the curve (5π‘₯/2𝑦) βˆ’ (2𝑦/π‘₯) = βˆ’4 at the point (2, 5).

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Video Transcript

Find the slope of the tangent to the curve five π‘₯ over two 𝑦 minus two 𝑦 over π‘₯ equals negative four at the point two, five.

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. Now, we’re most used to evaluating derivatives of explicitly defined functions of π‘₯. In this case though, we’re looking at a function that is defined implicitly. And so, we’re going to use implicit differentiation to evaluate the derivative.

This is just a special case of the chain rule. And it says that the derivative of some function in 𝑦 with respect to π‘₯ is equal to the derivative of that function with respect to 𝑦 times d𝑦 by dπ‘₯. And so, let’s look at the equation of our curve. It’s five π‘₯ over two 𝑦 minus two 𝑦 over π‘₯ equals negative four. And we’ll just begin by differentiating each side of this equation with respect to π‘₯.

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. And so, we’re going to differentiate five π‘₯ over two 𝑦, two 𝑦 over π‘₯, and negative four. Well, in fact, the derivative of a constant is zero. So, the right-hand side here becomes zero. But how do we differentiate five π‘₯ over two 𝑦 and two 𝑦 over π‘₯?

Well, we use the quotient rule. This says that the derivative of the quotient of two differentiable functions 𝑒 and 𝑣 is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. Let’s begin by looking at the derivative of five π‘₯ over two 𝑦. We’ll let 𝑒 be equal to five π‘₯, that’s the numerator, and 𝑣 be equal to two 𝑦, that’s the denominator. Then, d𝑒 by dπ‘₯, the first derivative of five π‘₯, is simply five. But what about d𝑣 by dπ‘₯?

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 𝑦, so that’s two, by d𝑦 by dπ‘₯. And so, d𝑣 by dπ‘₯ is simply two d𝑦 by dπ‘₯. Then, the derivative of five π‘₯ over two 𝑦 is 𝑣 times d𝑒 by dπ‘₯, so that’s two 𝑦 times five, minus 𝑒 times d𝑣 by dπ‘₯, so that’s five π‘₯ times two d𝑦 by dπ‘₯, all over 𝑣 squared, so two 𝑦 squared. If we simplify a little, we get 10𝑦 minus 10π‘₯ d𝑦 by dπ‘₯ all over four 𝑦 squared. And then, we should see that we can divide through by a common factor of two. And we found the derivative of the first part of our expression on the left-hand side. It’s five 𝑦 minus five π‘₯ d𝑦 by dπ‘₯ all over 𝑦 squared.

We’re now going to repeat this process for the derivative of two 𝑦 over π‘₯. This time, we let 𝑒 be equal to two 𝑦 and 𝑣 be equal to π‘₯. Of course, we know that the derivative of π‘₯ with respect to π‘₯ is just one. We already evaluated the derivative of two 𝑦. But as a reminder, it’s the derivative of two 𝑦 with respect to 𝑦 which is two times d𝑦 by dπ‘₯.

Let’s substitute everything we have into our formula for the quotient rule. And we get two π‘₯ d𝑦 by dπ‘₯ minus two 𝑦 over π‘₯ squared. And so, we find that when we differentiate both sides of our equation with respect to π‘₯, we get five 𝑦 minus five π‘₯ d𝑦 by dπ‘₯ over 𝑦 squared minus two π‘₯ d𝑦 by dπ‘₯ minus two 𝑦 over π‘₯ squared equals zero. But what on earth are we going to do with this?

Remember, we’re trying to find the slope of the tangent, so we need to evaluate d𝑦 by dπ‘₯ at our point two, five. The problem is, we haven’t currently got an expression for d𝑦 by dπ‘₯. So, let’s see if we can find a way to simplify a little. With our two fractions, we’re going to divide each bit of the numerator by the denominator. So, we divide five 𝑦 by 𝑦 squared to get five over 𝑦. And we divide negative five π‘₯ d𝑦 by dπ‘₯ by 𝑦 squared to get negative five π‘₯ over 𝑦 squared d𝑦 by dπ‘₯.

Similarly, we divide negative two π‘₯ d𝑦 by dπ‘₯ by π‘₯ squared to get negative two over π‘₯ d𝑦 by dπ‘₯ and negative negative two 𝑦 by π‘₯ squared to get positive two 𝑦 over π‘₯ squared. We now see that we can factor d𝑦 by dπ‘₯. And with those terms, we get d𝑦 by dπ‘₯ times negative five π‘₯ over 𝑦 squared minus two over π‘₯. At the same time, let’s subtract five over 𝑦 and two 𝑦 over π‘₯ squared from both sides. And so, we have this given expression.

We now should be able to see that we can divide through by negative five π‘₯ over 𝑦 squared minus two over π‘₯. We might also notice that every single term has a factor of negative one. So, we’re going to divide through by negative one at the same time. And so, we have a rather nasty-looking expression for the first derivative of 𝑦 with respect to π‘₯. It’s five over 𝑦 plus two 𝑦 over π‘₯ squared all over five π‘₯ over 𝑦 squared plus two over π‘₯.

Normally, we might look to simplify this a little, but remember we’re trying to find the slope of the tangent at a given point. In fact, it’s the point where π‘₯ equals two and 𝑦 equals five. So, let’s substitute π‘₯ equals two and 𝑦 equals five into our expression for the derivative.

The numerator becomes five over five plus two times five over two squared, which simplifies to one plus five over two. Then, our denominator is five times two over five squared plus two over two, which is two-fifths plus one. One plus five over two is seven over two. And two-fifths plus one is seven-fifths. So, we have seven over two divided by seven-fifths.

And then, we might recall that to divide by a fraction, we multiply the reciprocal of that same fraction. So, it’s seven over two times five-sevenths. And then the sevens cancel, to get five over two. And so, we’re done. We found the slope of the tangent to our curve at the point two, five. It’s five over two, or five-halves.

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