Given that 𝑥 equals five 𝑡 minus four times the natural log of 𝑡 and 𝑦 equals four 𝑡 plus five times the natural log of three 𝑡, find d𝑦 by d𝑥.
Here we’ve been given a pair of parametric equations. These are equations for 𝑥 and 𝑦 in terms of a third parameter. That’s 𝑡. And we’re looking to find the derivative of 𝑦 with respect to 𝑥. And so, we recall that as long as d𝑥 by d𝑡 is not equal to zero, the derivative of 𝑦 with respect to 𝑥 is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡. And so, it should be quite clear to us that we’re going to need to differentiate each of our respective equations with respect to 𝑡. And we’ll do this term by term.
The first term in our equation for 𝑥 is five 𝑡. Now, the derivative of five 𝑡 with respect to 𝑡 is five. Next, we recall that the derivative of the natural log of 𝑎𝑥, where 𝑎 is some constant, with respect to 𝑥 is one over 𝑥. So, the derivative of the natural log of 𝑡 is one over 𝑡. And so, the derivative of negative four times the natural log of 𝑡 is negative four over 𝑡.
Let’s repeat this process for d𝑦 by d𝑡. The derivative of four 𝑡 with respect to 𝑡 is four. Then, the derivative of the natural log of three 𝑡 must be one over 𝑡. So, five times the natural log of three 𝑡 is five over 𝑡. d𝑦 by d𝑥 is the quotient of these. It’s d𝑦 by d𝑡 divided by d𝑥 by d𝑡. So, that’s four plus five over 𝑡 divided by five minus four over 𝑡.
Now, at the moment, this doesn’t look particularly nice, so we’re going to deal with the fractions in our numerator and denominator. We’re going to multiply through by 𝑡. Then, our numerator becomes four 𝑡 plus five, and our denominator becomes five 𝑡 minus four. And so, we’re done. We found d𝑦 by d𝑥. It’s four 𝑡 plus five over five 𝑡 minus four.