Video: Finding the Second Derivative of Root Functions

Given that 𝑦 = √(π‘₯ βˆ’ 1), determine (d²𝑦)/(dπ‘₯Β²).

03:18

Video Transcript

Given that 𝑦 is equal to the square root of π‘₯ minus one, determine d two 𝑦 by dπ‘₯ squared.

We’re given 𝑦 as a function in π‘₯, and we’re asked to determine an expression for d two 𝑦 by dπ‘₯ squared. That’s the second derivative of 𝑦 with respect to π‘₯. So we’re going to need to differentiate our expression for 𝑦 twice. We can see 𝑦 is given as the composition of two functions. So we could do this by using the chain rule. And this would work. However, by using our laws of exponents, we can rewrite our expression for 𝑦 as π‘₯ minus one all raised to the power of one-half. Now our outer function is a power function, and we can differentiate this by using the general power rule. Either method would work; it’s personal preference which one you would use.

In this video, we’ll use the general power rule. We recall this tells us for any real constant 𝑛 and differentiable function 𝑓 of π‘₯, 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‘₯ all raised to the power of 𝑛 minus one. And we can see that our exponent 𝑛 is equal to one-half and our inner function 𝑓 of π‘₯ is the linear function π‘₯ minus one. To use the general power rule, we see we need to find an expression for 𝑓 prime of π‘₯. Since 𝑓 of π‘₯ is a linear function, its derivative with respect to π‘₯ will be the coefficient of π‘₯, which is one.

And now that we found an expression for 𝑓 prime, we can use the general power rule to find an expression for d𝑦 by dπ‘₯. We substitute 𝑛 is equal to one-half, 𝑓 of π‘₯ is equal to π‘₯ minus one, and 𝑓 prime of π‘₯ is equal to one into our general power rule. This gives us d𝑦 by dπ‘₯ is equal to one-half times one multiplied by π‘₯ minus one all raised to the power of one-half minus one. And we can simplify our coefficient and our exponent to get d𝑦 by dπ‘₯ is equal to one-half times π‘₯ minus one all raised to the power of negative one-half. And we could simplify this by using our laws of exponents. But remember, we’re trying to find an expression for d two 𝑦 by dπ‘₯ squared. So we’re going to need to differentiate d𝑦 by dπ‘₯ with respect to π‘₯.

And in this current form, we can differentiate this once again by using our general power rule. This time, we can see our exponent 𝑛 is equal to negative one-half and our inner function 𝑓 of π‘₯ is still equal to π‘₯ minus one. So we can use the general power rule to find an expression for our second derivative of 𝑦 with respect to π‘₯. This time our value of 𝑛 is negative one-half, 𝑓 prime is still equal to one, and 𝑓 of π‘₯ is equal to π‘₯ minus one. And remember, since we’re multiplying our expression by one-half, we need to multiply our derivative by one-half. This gives us one-half times negative one-half multiplied by one times π‘₯ minus one to the power of negative one-half minus one.

And just as we did before, we can simplify our coefficient and our exponent. This gives us d two 𝑦 by dπ‘₯ squared is equal to negative one-quarter times π‘₯ minus one all raised to the power of negative three over two. And we could leave our answer like this. However, we can also simplify this by using our laws of exponents. First, remember, raising a number to a negative exponent is the same as dividing by that number raised to the positive exponent. So, π‘Ž to the power of negative three over two is equal to one divided by π‘Ž to the power of three over two. We can also remember that raising a number to the power of three over two is the same as cubing that number and taking the square root.

We can use this to rewrite our answer. We’ll set our value of π‘Ž equal to π‘₯ minus one. And this gives us our final answer. Therefore, we were able to show if 𝑦 is equal to the square root of π‘₯ minus one, then the second derivative of 𝑦 with respect to π‘₯ is equal to negative one divided by four times the square root of π‘₯ minus one all cubed.

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