Find the first derivative with respect to x
One formula for every slope
A point moves with \(x = 2e^{t}\) and \(y = e^{2t}\). The upper graph shows its path, and the lower graph shows the slope of the path against \(t\). The orange bar is as high as the slope at the orange point.
The legs of the small triangle are the two rates, \(\frac{dx}{dt}\) across and \(\frac{dy}{dt}\) up, and they are green because both rates are positive. At \(t = 0\), the point is at \((2, 1)\) and both rates are 2, so the slope is 1.
Move \(t\) to \(\ln 2\). The rates are 4 and 8, so the slope is 2, and the bar is 2 high.
For any \(t\), the ratio is \(\frac{2e^{2t}}{2e^{t}} = e^{t}\), the formula that the lower graph shows. This one formula gives the slope at every point, so you do not divide again for each \(t\).
The slope as a function of t
The same scene has no numbers, and the legs of the triangle are labelled with the two rates.
Move \(t\). At every point, the orange bar is as high as the vertical leg divided by the horizontal leg.
The lower graph plots this ratio, the first derivative \(\frac{dy}{dx}\), against the parameter \(t\), so its input is \(t\) and not \(x\).
Notation
For a curve \(x = f(t)\), \(y = g(t)\), differentiate \(x\) and \(y\) separately with respect to \(t\). The results are the two rates, the legs of the triangle.
\[\dfrac{dx}{dt} = f'(t),\qquad \dfrac{dy}{dt} = g'(t)\]Divide the rate of \(y\) by the rate of \(x\), because the slope is the rise over the run. The orange bar on the lower graph is this slope.
\[\cy{\dfrac{dy}{dx}} = \dfrac{\cg{dy/dt}}{\cg{dx/dt}}\]Simplify. If the top or the bottom is a fraction, multiply both by the same quantity to remove it. Cancel a factor only if it divides the whole top and the whole bottom.
Keep the answer as a function of \(t\), and substitute a value of \(t\) only when the question gives one.
The formula needs \(x\) and \(y\) to be differentiable in \(t\), and it needs \(\frac{dx}{dt} \ne 0\), because it divides by that rate.
The first derivative equals the rate of y divided by the rate of x.
When it fails
On the curve \(x = t^3 - 3t\), \(y = t\), the slope is \(\frac{1}{3t^2 - 3}\). The bottom is 0 at \(t = 1\) and at \(t = -1\). Between them \(\frac{dx}{dt} < 0\), so the horizontal leg is red and points left.
Move \(t\) to 1. There \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 1\), so the point moves straight up and the tangent line is vertical. The formula has no value at \(t = 1\), so the lower graph breaks.
Example 1: A rate that is a fraction
For \(x = \ln t\) and \(y = t^2 - 1\), with \(t > 0\), find \(\frac{dy}{dx}\) as a function of \(t\). Then find the slope at \(t = 1\).
Differentiate \(x\) and \(y\) with respect to \(t\).
\[\dfrac{dx}{dt} = \dfrac1t,\qquad \dfrac{dy}{dt} = 2t\]Divide. Dividing by \(\frac1t\) is the same as multiplying by \(t\).
\[\cy{\dfrac{dy}{dx}} = \dfrac{2t}{1/t} = 2t^2\]The bottom \(\frac1t\) is not 0 for any \(t > 0\), so the formula holds for all of them. At \(t = 1\) the legs are 1 and 2, so the slope is 2.
\[\cy{\dfrac{dy}{dx}}\Big|_{t=1} = 2\]To test the result, eliminate \(t\). Because \(t = e^{x}\), the curve is \(y = e^{2x} - 1\), and its derivative is \(2e^{2x} = 2t^2\), the same result.
Example 2: A formula that simplifies
For \(x = \sin t\) and \(y = \cos 2t\), with \(-\frac{\pi}{2} < t < \frac{\pi}{2}\), find \(\frac{dy}{dx}\) as a function of \(t\).
Differentiate \(x\) and \(y\) with respect to \(t\). The second one needs the chain rule.
\[\dfrac{dx}{dt} = \cos t,\qquad \dfrac{dy}{dt} = -2\sin 2t\]A tempting wrong move: Divide the rate of \(x\) by the rate of \(y\).
\[\dfrac{dy}{dx}\ \overset{?}{=}\ \cr{\dfrac{\cos t}{-2\sin 2t}}\]At \(t = \frac{\pi}{6}\), this gives \(-\frac12\). The red line with this slope is much flatter than the curve at the orange point. The slope is the rise over the run, so the rate of \(y\) goes on top.
Put the rate of \(y\) on top, and use \(\sin 2t = 2\sin t\cos t\) to simplify.
\[\cy{\dfrac{dy}{dx}} = \dfrac{-2\sin 2t}{\cos t} = \dfrac{-4\sin t\cos t}{\cos t}\]The factor \(\cos t\) is not 0 for \(-\frac{\pi}{2} < t < \frac{\pi}{2}\), so it cancels. At \(t = \frac{\pi}{6}\) the slope is \(-2\).
\[\cy{\dfrac{dy}{dx}} = -4\sin t\]To test the result, use \(\cos 2t = 1 - 2\sin^2 t\). Then \(y = 1 - 2x^2\), and its derivative is \(-4x = -4\sin t\), the same result.
Practice
For \(x = t + \dfrac1t\) and \(y = \ln t\), with \(t > 1\), find \(\frac{dy}{dx}\) as a function of \(t\). Then find the slope at \(t = 2\).
Write \(\frac{dx}{dt}\) as one fraction, because dividing by a fraction is multiplying by its reciprocal. The orange bar shows the slope to compare with.
Solution
Differentiate. Write \(\frac{dx}{dt}\) as one fraction.
\[\begin{aligned} \dfrac{dx}{dt} &= 1 - \dfrac{1}{t^2} = \dfrac{t^2 - 1}{t^2} \\ \dfrac{dy}{dt} &= \dfrac1t \end{aligned}\]Divide. Multiply by the reciprocal of the bottom, and cancel one factor \(t\).
\[\cy{\dfrac{dy}{dx}} = \dfrac1t\cdot\dfrac{t^2}{t^2 - 1} = \dfrac{t}{t^2 - 1}\]The bottom \(t^2 - 1\) is not 0 for \(t > 1\). At \(t = 2\) the slope is \(\frac{2}{4 - 1} = \frac23\).
\[\cy{\dfrac{dy}{dx}}\Big|_{t=2} = \dfrac23\]
Questions
The upper graph shows the curve \(x = t^3\), \(y = 3t^2\), for \(t > 0\). The lower graph shows its slope against \(t\). The orange point and the orange bar are at the same value of \(t\).
What is the slope of the curve at the orange point, where t = 1?
- 2
- \(\frac12\)
- 6
Answer
2
As t gets closer to 0 from the right, what happens to the slope?
- It grows without bound
- It approaches 0
- It stays equal to 2
Answer
It grows without bound
For \(x = \dfrac{1}{t + 1}\) and \(y = t^2\), what is \(\frac{dy}{dx}\) at \(t = 1\)?
- −8
- \(-\frac18\)
- 8
Answer
−8
Summary
You differentiate \(x\) and \(y\) with respect to \(t\), put the rate of \(y\) over the rate of \(x\), and simplify.
It works because the slope is the rise over the run, and it gives one formula in \(t\) for the slope at every point.
You use it for a curve given by \(x = f(t)\) and \(y = g(t)\), at values of \(t\) where \(\frac{dx}{dt}\) is not 0.
This is one of four skills in topic 9.2 of Which & Why: AP Calculus.