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

Given that π‘₯ = 3𝑒^(5𝑑) and 𝑦 = 𝑑𝑒^(βˆ’5𝑑), find d𝑦/dπ‘₯.

03:08

Video Transcript

Given that π‘₯ equals three 𝑒 to the power of five 𝑑 and 𝑦 equals 𝑑 times 𝑒 to the power of negative five 𝑑, find d𝑦 by dπ‘₯.

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

We’ll begin by differentiating π‘₯ with respect to 𝑑. Now, π‘₯ is equal to three 𝑒 to the power of five 𝑑. And we know that to differentiate 𝑒 to the power of π‘Žπ‘₯ for some constant value π‘Ž, we get π‘Ž times 𝑒 to the power of π‘Žπ‘₯. Well, in this case, we’re differentiating three 𝑒 to the power of five 𝑑. So we multiply the entire term by the coefficient of 𝑑, which is five. And we see that dπ‘₯ by d𝑑 is five times three 𝑒 to the power of five 𝑑, which is 15𝑒 to the power of five 𝑑.

Next, we need to differentiate 𝑦 with respect to 𝑑. But we see that 𝑦 itself is the product of two differentiable functions. It’s 𝑑 times 𝑒 to the power of negative five 𝑑. And so we’re going to use the product rule. This says that the derivative of the product of two differentiable functions 𝑒 and 𝑣 is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. Well, our two functions are 𝑑 and 𝑒 to the power of negative five 𝑑. So d𝑦 by d𝑑 is 𝑑 times the derivative of 𝑒 to the power of negative five 𝑑 with respect to 𝑑 plus 𝑒 to the power of negative five 𝑑 times the derivative of 𝑑 with respect to 𝑑.

Now, the derivative of 𝑒 to the power of negative five 𝑑 we know to be equal to negative five 𝑒 to the power of negative five 𝑑. And the derivative of 𝑑 with respect to 𝑑 is simply one. So d𝑦 by d𝑑 is 𝑑 times negative five 𝑒 to the power of negative five 𝑑 plus 𝑒 to the power of negative five 𝑑, which simplifies as shown. d𝑦 by dπ‘₯ is the quotient of these. It’s d𝑦 by d𝑑 divided by dπ‘₯ by d𝑑. So that’s negative five 𝑑𝑒 to the power of negative five 𝑑 plus 𝑒 to the power of negative five 𝑑 all over 15𝑒 to the power five 𝑑.

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

And we therefore find d𝑦 by dπ‘₯ to be equal to one minus five 𝑑 over 15𝑒 to the power of 10𝑑.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.