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Given π¦ equals the square root of π₯ minus nine, find d two π¦ by dπ₯ squared.
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This notation, which I read as d two π¦ by dπ₯ squared, means the second derivative of π¦ with respect to π₯.
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We differentiate π¦ with respect to π₯ once to find dπ¦ by dπ₯.
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And then, we different with respect to π₯ a second time to give the second derivative of π¦ with respect to π₯.
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So, weβre going to differentiate twice.
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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.
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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.
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We multiply by the power, reduce the power by one.
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But we also multiply by the derivative of the function π of π₯.
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Here, our function π of π₯ is π₯ minus nine.
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And its derivative with respect to π₯, π prime of π₯, is just one, which simplifies things somewhat.
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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.
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Thatβs one-half minus one.
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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.
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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.
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We can apply the general power rule a second time.
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The factor of one-half just acts as a multiplicative constant.
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So we have one-half multiplied by negative a half, thatβs the previous power.
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Multiplied by one, thatβs the derivative of π₯ minus nine, our function π of π₯.
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Multiplied by π₯ minus nine to the power of negative three over two.
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Weβve decreased the power by one.
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That simplifies to negative one over four π₯ minus nine to the power of three over two.
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If we recall that a negative power defines a reciprocal.
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So by differentiating π¦ first once and then a second time with respect to π₯.
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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.