# Video: Finding the First Derivative of a Function Defined by Parametric Equations Involving Using the Product Rule

Given that 𝑥 = 3𝑒^(5𝑡) and 𝑦 = 𝑡𝑒^(−5𝑡), find d𝑦/d𝑥.

03:08

### Video Transcript

Given that 𝑥 equals three 𝑒 to the power of five 𝑡 and 𝑦 equals 𝑡 times 𝑒 to the power of negative five 𝑡, find d𝑦 by d𝑥.

In this question, we’ve been given a pair of parametric equations. These are equations for 𝑥 and 𝑦 in terms of a third parameter; here, that’s 𝑡. We’re asked to differentiate 𝑦 with respect to 𝑥. And so we recall that if d𝑦 by d𝑡 is not equal to zero, d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡. And so we can see that we’re going to need to differentiate the equation for 𝑥 with respect to 𝑡 and the equation for 𝑦 with respect to 𝑡.

We’ll begin by differentiating 𝑥 with respect to 𝑡. Now, 𝑥 is equal to three 𝑒 to the power of five 𝑡. And we know that to differentiate 𝑒 to the power of 𝑎𝑥 for some constant value 𝑎, we get 𝑎 times 𝑒 to the power of 𝑎𝑥. Well, in this case, we’re differentiating three 𝑒 to the power of five 𝑡. So we multiply the entire term by the coefficient of 𝑡, which is five. And we see that d𝑥 by d𝑡 is five times three 𝑒 to the power of five 𝑡, which is 15𝑒 to the power of five 𝑡.

Next, we need to differentiate 𝑦 with respect to 𝑡. But we see that 𝑦 itself is the product of two differentiable functions. It’s 𝑡 times 𝑒 to the power of negative five 𝑡. And so we’re going to use the product rule. This says that the derivative of the product of two differentiable functions 𝑢 and 𝑣 is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. Well, our two functions are 𝑡 and 𝑒 to the power of negative five 𝑡. So d𝑦 by d𝑡 is 𝑡 times the derivative of 𝑒 to the power of negative five 𝑡 with respect to 𝑡 plus 𝑒 to the power of negative five 𝑡 times the derivative of 𝑡 with respect to 𝑡.

Now, the derivative of 𝑒 to the power of negative five 𝑡 we know to be equal to negative five 𝑒 to the power of negative five 𝑡. And the derivative of 𝑡 with respect to 𝑡 is simply one. So d𝑦 by d𝑡 is 𝑡 times negative five 𝑒 to the power of negative five 𝑡 plus 𝑒 to the power of negative five 𝑡, which simplifies as shown. d𝑦 by d𝑥 is the quotient of these. It’s d𝑦 by d𝑡 divided by d𝑥 by d𝑡. So that’s negative five 𝑡𝑒 to the power of negative five 𝑡 plus 𝑒 to the power of negative five 𝑡 all over 15𝑒 to the power five 𝑡.

By recognizing that 𝑒 to the power of negative five 𝑡 is the same as one over 𝑒 to the power of five 𝑡 and then factoring by one over 𝑒 to the power of five 𝑡. We see we can write this as one over 𝑒 to the power of five 𝑡 times negative five 𝑡 plus one over 15𝑒 to the power of five 𝑡. But of course, we would generally write this second fraction as one minus five 𝑡 over 15𝑒 to the power of five 𝑡. Now, when we multiply two numbers with exponents, we add those exponents. So 𝑒 to the power of five 𝑡 times 15𝑒 to the power of five 𝑡 is 15𝑒 to the power of 10𝑡.

And we therefore find d𝑦 by d𝑥 to be equal to one minus five 𝑡 over 15𝑒 to the power of 10𝑡.