Video Transcript
Given 𝑦 equals the square root of 𝑥 minus nine, find d two 𝑦 by d𝑥 squared.
This notation, which I read as d two 𝑦 by d𝑥 squared, means the second derivative of 𝑦 with respect to 𝑥. We differentiate 𝑦 with respect to 𝑥 once to find d𝑦 by d𝑥. And then, we different with respect to 𝑥 a second time to give the second derivative of 𝑦 with respect to 𝑥. 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 𝑥 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 𝑥 to the power of 𝑛, then its derivative with respect to 𝑥 is equal to 𝑛 multiplied by 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 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 𝑥.
Here, our function 𝑓 of 𝑥 is 𝑥 minus nine. And its derivative with respect to 𝑥, 𝑓 prime of 𝑥, is just one, which simplifies things somewhat. Applying the general power rule then, we see that the first derivative of 𝑦 with respect to 𝑥, d𝑦 by d𝑥, is equal to a half multiplied by one multiplied by 𝑥 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 𝑥 minus nine to the power of negative a half.
To find the second derivative of 𝑦 with respect to 𝑥, d two 𝑦 by d𝑥 squared, we take the derivative we’ve just found and we differentiate with respect to 𝑥 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 𝑥 minus nine, our function 𝑓 of 𝑥. Multiplied by 𝑥 minus nine to the power of negative three over two. We’ve decreased the power by one. That simplifies to negative one over four 𝑥 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 𝑥. We’ve found that the second derivative of 𝑦 with respect to 𝑥, d two 𝑦 by d𝑥 squared, is equal to negative one over four 𝑥 minus nine to the power of three over two.
