Video: Finding the Third Derivative of Root Functions

Given that 𝑦 = √(2π‘₯ βˆ’ 5), determine 𝑦‴.

04:25

Video Transcript

Given that 𝑦 is equal to the square root of two π‘₯ minus five, determine 𝑦 triple prime.

We’re given 𝑦 as a function in π‘₯, and we’re asked to determine 𝑦 triple prime. That’s the third derivative of 𝑦 with respect to π‘₯. So we’re going to need to differentiate this expression for 𝑦 three times. We can see 𝑦 is given as the composition of two functions. So one way of doing this would be to use the chain rule. And this would work. However, because our outer function is the square root, we can also do this by using the general power rule.

First, by using our laws of exponents, we can rewrite 𝑦 as two π‘₯ minus five all raised to the power of one-half. And now our outer function is a power function. So we can differentiate this by using the general power rule. So we’ll start by recalling the general power rule. This tells us for any differentiable function 𝑓 of π‘₯ and real constant 𝑛, the derivative of 𝑓 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ raised to the power of 𝑛 minus one. In our case, we can see our exponent 𝑛 is equal to one-half and our inner function is the linear function two π‘₯ minus five.

And now we can see, to use the general power rule, we need to find an expression for 𝑓 prime of π‘₯. That’s the derivative of two π‘₯ minus five with respect to π‘₯. Of course this is a linear function, so its derivative will be the coefficient of π‘₯, which is two. Now, we can substitute 𝑛 is equal to one-half, 𝑓 of π‘₯ is two π‘₯ minus five, and 𝑓 prime of π‘₯ is two into our general power rule. This gives us 𝑦 prime is equal to one-half times two multiplied by two π‘₯ minus five all raised to the power of one-half minus one.

And of course we can simplify this. First, one-half times two is equal to one. Next, in our exponent, one-half minus one is equal to negative one-half. So we’ve shown 𝑦 prime is equal to two π‘₯ minus five all raised to the power of negative one-half. Of course we could simplify this further. However, we need to differentiate this two more times. And we’re going to do this again by using the general power rule.

This time, we can see our value of the exponent 𝑛 is equal to negative one-half. We can see our inner function 𝑓 of π‘₯ stays the same, so 𝑓 prime of π‘₯ will also be the same. So the only difference is, instead of having 𝑛 as one-half, we’re going to have 𝑛 as negative one-half. So by applying the general power rule with 𝑛 equal to negative one-half, 𝑓 of π‘₯ equal to two π‘₯ minus five, and 𝑓 prime of π‘₯ equal to two, we get 𝑦 double prime is equal to negative one-half times two multiplied by two π‘₯ minus five all raised to the power of negative one-half minus one.

And once again we can simplify. Again, we have negative one-half multiplied by two. This gives us negative one. And in our exponent, we have negative one-half minus one, which is of course equal to negative three over two. Therefore, we’ve shown 𝑦 double prime is equal to negative one times two π‘₯ minus five all raised to the power of negative three over two.

And again, we could simplify this by using our laws of exponents. But we still need to differentiate this one more time. And we can see we can do this by using the general power rule. This time, our exponent will be equal to negative three over two. And we can see our inner function 𝑓 of π‘₯ is the same. So 𝑓 prime of π‘₯ will also be the same. The only difference is our exponent is negative three over two and we’re multiplying our entire function by negative one.

So we’ll differentiate this one more time by using the general power rule. We set our exponent 𝑛 equal to negative three over two, 𝑓 of π‘₯ to be two π‘₯ minus five, and 𝑓 prime of π‘₯ as once again equal to two. And remember, we have a coefficient of negative one. This gives us 𝑦 triple prime is equal to negative one times negative three over two multiplied by two times two π‘₯ minus five all raised to the power of negative three over two minus one.

And we can simplify this. First, negative one multiplied by negative three over two multiplied by two is equal to three. And in our exponent, negative three over two minus one is equal to negative five over two. Therefore, we’ve shown 𝑦 triple prime is equal to three times two π‘₯ minus five all raised to the power of negative five over two.

And we could leave our answer like this. However, we don’t need to differentiate this anymore. So we can simplify this by using our laws of exponents. First, by using our laws of exponents, we know π‘Ž to the power of negative five over two will be equal to one divided by π‘Ž to the power of five over two. We could then simplify this even further by using our laws of exponents. We know π‘Ž to the power of five over two is equal to the square root of π‘Ž to the fifth power. And we can apply this to simplify 𝑦 triple prime. We set our value of π‘Ž equal to two π‘₯ minus five. And doing this gives us three divided by the square root of two π‘₯ minus five all raised to the fifth power, which is our final answer.

In this question, we were given that 𝑦 was equal to the square root of two π‘₯ minus five. And we were able to use the general power rule three times to show that 𝑦 triple prime was equal to three divided by the square root of two π‘₯ minus five all raised to the fifth power.

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