If 𝑥 equals negative eight 𝑡 to the fifth power minus eight and 𝑦 equals the fifth root of 𝑡 to the sixth power, find d𝑦 by d𝑥 at 𝑡 equals one.
These functions have been defined parametrically. This means that rather than 𝑦 being given explicitly as a function of 𝑥, both 𝑦 and 𝑥 are given as functions of a third parameter 𝑡. We’re going to find d𝑦 by d𝑥 at the point 𝑡 equals one. The chain rule tells us that d𝑦 by d𝑡 is equal to d𝑦 by d𝑥 multiplied by d𝑥 by d𝑡. And we can rearrange this to find that d𝑦 by d𝑥 equals d𝑦 by d𝑡 over d𝑥 by d𝑡. And this is what we’re going to use to find d𝑦 by d𝑥.
And of course, this only works providing that the denominator d𝑥 by d𝑡 is not equal to zero. So we need to find d𝑦 by d𝑡 and d𝑥 by d𝑡. Let’s begin by calculating d𝑦 by d𝑡. We have that 𝑦 equals the fifth root of 𝑡 to the sixth power. To find d𝑦 by d𝑡, we’re going to differentiate this with respect to 𝑡. This appears to be a fairly complicated function to differentiate. However, if we remember the fact that the 𝑚th root of 𝑥 to the 𝑛th power can be written as 𝑥 to the power of 𝑛 over 𝑚, we can see that we can rewrite 𝑦 as 𝑡 to the six-fifths power.
Then to differentiate this, we recall the power rule of differentiation. This tells us that the derivative of 𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑛 multiplied by 𝑥 to the 𝑛 minus one power. So essentially, we multiply by the power and then subtract one from the power. And applying this to the function 𝑦, we find that d𝑦 by d𝑡 is equal to six-fifths 𝑡 raised to the one-fifth power. We find d𝑥 by d𝑡 by differentiating 𝑥 equals negative 𝑡 to the fifth power minus eight with respect to 𝑡.
We can do this term by term starting with the term negative eight 𝑡 to the fifth power. Applying the power rule to negative eight 𝑡 to the fifth power, we find that d𝑥 by d𝑡 is equal to negative 40𝑡 to the fourth power. As for the negative eight, because this is just a constant, constants differentiate to zero. So d𝑥 by d𝑡 is just negative 40𝑡 to the fourth power. Now, remember that we said that d𝑥 by d𝑡 cannot be equal to zero. And because we’re evaluating d𝑦 by d𝑥 at the point 𝑡 equals one, at this point, d𝑥 by d𝑡 is not equal to zero. So this is okay.
Putting this together, we have that d𝑦 by d𝑥 equals d𝑦 by d𝑡, which is six-fifths 𝑡 to the one-fifth power over d𝑥 by d𝑡, which is negative 40𝑡 to the fourth power. Remember, we’re evaluating this at the point 𝑡 equals one. So we substitute 𝑡 equals one, and we remember that raising one to any power is always going to give us one. So both one to the one-fifth power and one to the fourth power are both going to give us one. So we’re left with d𝑦 by d𝑥 equals six-fifths over negative 40. And we can actually bring the negative on the denominator out to the front of the fraction.
Now, in order to simplify this, one way we could think of this fraction is as negative six-fifths divided by 40 or equivalently negative six-fifths divided by 40 over one. And we can divide these two fractions by changing the divide to a times and flipping the second fraction. This gives us negative six-fifths times one over 40, which is negative six over 200. We can then simplify this by canceling down to give us negative three over 100. And let’s then write this in decimal form. Of course, negative three divided by 100 is negative 0.03. So that gives us our final answer.