# Video: Finding the First Derivative of a Function Defined by Parametric Equations

Given that 𝑦 = −7𝑡³ + 8 and 𝑧 = −7𝑡² + 3, find the rate of change of 𝑦 with respect to 𝑧.

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### Video Transcript

Given that 𝑦 equals negative seven [𝑡] cubed plus eight and 𝑧 equals negative seven [𝑡] squared plus three, find the rate of change of 𝑦 with respect to 𝑧.

When faced with a question about the rate of change of something, we should be thinking about derivatives. Here we want to find the rate of change of 𝑦 with respect to 𝑧. So we’re going to work out d𝑦 by d𝑧. That’s the first derivative of 𝑦 with respect to 𝑧.

We then recall that, given two parametric equations — 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡 — we find d𝑦 by d𝑥 by multiplying d𝑦 by d𝑡 by one over d𝑥 by d𝑡. Or equivalently, by dividing d𝑦 by d𝑡 by d𝑥 by d𝑡. In this example, our two functions are 𝑦 and 𝑧. So we say that d𝑦 by d𝑧 equals d𝑦 by d𝑡 divided by d𝑧 by d𝑡. And we see that we’re going to need to begin by differentiating each function with respect to 𝑡.

We’ll begin by differentiating 𝑦 with respect to 𝑡. Remember, to differentiate a polynomial term, we multiply the term by the exponent and then reduce that exponent by one. So the first derivative of negative 70 cubed is three times negative 70 squared. And actually, the first derivative of eight is zero. Of course, we don’t really need to include that plus zero. So we find that d𝑦 by d𝑡 is equal to negative 21𝑡 squared.

We’ll now repeat this for d𝑧 by d𝑡. This time, the first derivative is two times negative seven 𝑡, which is negative 14𝑡. d𝑦 by d𝑧 is what we get when we divide d𝑦 by d𝑡 by d𝑧 by d𝑡. So that’s negative 21𝑡 squared divided by negative 14𝑡. Of course, a negative divided by a negative is a positive. And we can divide both the numerator and the denominator by 𝑡. Our final step is to simplify by dividing by 21 and 14 by seven. So we find the rate of change of 𝑦 with respect to 𝑧 to be three 𝑡 over two.