Video: Finding the Equation of the Tangent to the Curve of a Function at a Point

Find the equation of the tangent to the curve 𝑦² = π‘₯⁴ at the point (βˆ’1, 1).

03:24

Video Transcript

Find the equation of the tangent to the curve 𝑦 squared equals π‘₯ to the fourth power at the point negative one, one.

In this question, we’ve been asked to find the equation of a tangent to a curve. And so, we begin by recalling the equation of a straight line. For a line that passes through the point π‘₯ one, 𝑦 one with a slope of π‘š, its equation is 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. Now, we know that the point that the tangent passes through is negative one, one. But what’s its slope? Well, the slope of the tangent to a curve is found by differentiating the function and then evaluating it at the point negative one, one.

We will, however, need to be a little bit careful about how we differentiate our function. It might be tempting to take the square root of both sides of this equation, to get 𝑦 is equal to π‘₯ squared. The problem is, if we do that, we have to take both the positive and negative square root of π‘₯ to the fourth power. So, instead, we’re going to use a process called implicit differentiation. In implicit differentiation, we use the chain rule. We say that the derivative of a function in 𝑦 with respect to π‘₯ is equal to the derivative of that function with respect to 𝑦 times d𝑦 by dπ‘₯.

So, let’s take our equation 𝑦 squared equals π‘₯ to the fourth power. We want to differentiate both sides of this equation with respect to π‘₯. Well, the rule for implicit differentiation says that the derivative of 𝑦 squared with respect to π‘₯ is the derivative of 𝑦 squared with respect to 𝑦 times d𝑦 by dπ‘₯. Well, when we differentiate 𝑦 squared, we multiply the entire term by two and then reduce the exponent by one. So, we get two 𝑦. And we see that the left-hand side is then two 𝑦 d𝑦 by dπ‘₯.

The derivative of π‘₯ to the fourth power with respect to π‘₯ is four π‘₯ cubed. Once again, we multiply the entire term by four and then reduce that exponent by one to get three. If we then make d𝑦 by dπ‘₯ the subject, by dividing through by two 𝑦, we find that d𝑦 by dπ‘₯ is equal to four π‘₯ cubed over two 𝑦. And we now have an expression for the derivative in terms of π‘₯ and 𝑦.

Of course, we said that to find the equation of the tangent, we need to know the slope of the tangent. And we can find the slope by evaluating the derivative at the point we’re interested in. So here, that’s negative one, one. Now, of course, negative one, one is the point whose π‘₯-coordinate is negative one and whose 𝑦-coordinate is one. And so, we find that the derivative at the point negative one, one is four times negative one cubed over two times one. And we see that the slope π‘š is, therefore, negative two.

We now know the slope of the tangent to the curve and a point which it passes through. So, let’s substitute everything we have into our equation for a straight line. When we do, we get 𝑦 minus one equals negative two times π‘₯ minus negative one. π‘₯ minus negative one is π‘₯ plus one. And then, we distribute the parentheses on the right-hand side by multiplying each term by negative two. And so, we find 𝑦 minus one is equal to negative two π‘₯ minus two.

Our final step is to rearrange by adding one to both sides. And when we do, we realize we found the equation of the tangent to the curve at the point negative one, one. It’s 𝑦 equals negative two π‘₯ minus one.

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