Video: Finding the Second Derivative of Root Functions

Given 𝑦 = √(π‘₯ βˆ’ 9), find d²𝑦/dπ‘₯Β².

02:44

Video Transcript

Given 𝑦 equals the square root of π‘₯ minus nine, find d two 𝑦 by dπ‘₯ squared.

This notation, which I read as d two 𝑦 by dπ‘₯ squared, means the second derivative of 𝑦 with respect to π‘₯. We differentiate 𝑦 with respect to π‘₯ once to find d𝑦 by dπ‘₯. And then, we different with respect to π‘₯ a second time to give the second derivative of 𝑦 with respect to π‘₯. So, we’re going to differentiate twice.

Before we can do this though, we need to express 𝑦 using index notation rather than a surd . So we can write 𝑦 equals π‘₯ minus nine to the power of one-half. In order to differentiate, we can recall the general power rule, which tells us that if 𝑦 is some function 𝑓 of π‘₯ to the power of 𝑛, then its derivative with respect to π‘₯ is equal to 𝑛 multiplied by 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ to the power of 𝑛 minus one. We multiply by the power, reduce the power by one. But we also multiply by the derivative of the function 𝑓 of π‘₯.

Here, our function 𝑓 of π‘₯ is π‘₯ minus nine. And its derivative with respect to π‘₯, 𝑓 prime of π‘₯, is just one, which simplifies things somewhat. Applying the general power rule then, we see that the first derivative of 𝑦 with respect to π‘₯, d𝑦 by dπ‘₯, is equal to a half multiplied by one multiplied by π‘₯ minus nine to the power of negative a half. That’s one-half minus one. Of course, multiplying by one has no effect, so we can simplify to a half π‘₯ minus nine to the power of negative a half.

To find the second derivative of 𝑦 with respect to π‘₯, d two 𝑦 by dπ‘₯ squared, we take the derivative we’ve just found and we differentiate with respect to π‘₯ again. We can apply the general power rule a second time. The factor of one-half just acts as a multiplicative constant. So we have one-half multiplied by negative a half, that’s the previous power. Multiplied by one, that’s the derivative of π‘₯ minus nine, our function 𝑓 of π‘₯. Multiplied by π‘₯ minus nine to the power of negative three over two. We’ve decreased the power by one. That simplifies to negative one over four π‘₯ minus nine to the power of three over two. If we recall that a negative power defines a reciprocal.

So by differentiating 𝑦 first once and then a second time with respect to π‘₯. We’ve found that the second derivative of 𝑦 with respect to π‘₯, d two 𝑦 by dπ‘₯ squared, is equal to negative one over four π‘₯ minus nine to the power of three over two.

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