### Video Transcript

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

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

Before we can do this though, we need to express π¦ using index notation rather than a surd . So we can write π¦ 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 π¦ is some function π of π₯ to the power of π, then its derivative with respect to π₯ is equal to π multiplied by π prime of π₯ multiplied by π of π₯ to the power of π minus one. We multiply by the power, reduce the power by one. But we also multiply by the derivative of the function π of π₯.

Here, our function π of π₯ is π₯ minus nine. And its derivative with respect to π₯, π prime of π₯, is just one, which simplifies things somewhat. Applying the general power rule then, we see that the first derivative of π¦ with respect to π₯, dπ¦ by dπ₯, is equal to a half multiplied by one multiplied by π₯ minus nine to the power of negative a half. Thatβs 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 π¦ with respect to π₯, d two π¦ by dπ₯ squared, we take the derivative weβve 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, thatβs the previous power. Multiplied by one, thatβs the derivative of π₯ minus nine, our function π of π₯. Multiplied by π₯ minus nine to the power of negative three over two. Weβve 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 π¦ first once and then a second time with respect to π₯. Weβve found that the second derivative of π¦ with respect to π₯, d two π¦ by dπ₯ squared, is equal to negative one over four π₯ minus nine to the power of three over two.