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Given that π¦ is equal to the square root of π₯ minus one, determine d two π¦ by dπ₯ squared.
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Weβre given π¦ as a function in π₯, and weβre asked to determine an expression for d two π¦ by dπ₯ squared.
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Thatβs the second derivative of π¦ with respect to π₯.
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So weβre going to need to differentiate our expression for π¦ twice.
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We can see π¦ is given as the composition of two functions.
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So we could do this by using the chain rule.
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And this would work.
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However, by using our laws of exponents, we can rewrite our expression for π¦ as π₯ minus one all raised to the power of one-half.
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Now our outer function is a power function, and we can differentiate this by using the general power rule.
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Either method would work; itβs personal preference which one you would use.
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In this video, weβll use the general power rule.
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We recall this tells us for any real constant π and differentiable function π of π₯, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one.
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And we can see that our exponent π is equal to one-half and our inner function π of π₯ is the linear function π₯ minus one.
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To use the general power rule, we see we need to find an expression for π prime of π₯.
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Since π of π₯ is a linear function, its derivative with respect to π₯ will be the coefficient of π₯, which is one.
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And now that we found an expression for π prime, we can use the general power rule to find an expression for dπ¦ by dπ₯.
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We substitute π is equal to one-half, π of π₯ is equal to π₯ minus one, and π prime of π₯ is equal to one into our general power rule.
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This gives us dπ¦ by dπ₯ is equal to one-half times one multiplied by π₯ minus one all raised to the power of one-half minus one.
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And we can simplify our coefficient and our exponent to get dπ¦ by dπ₯ is equal to one-half times π₯ minus one all raised to the power of negative one-half.
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And we could simplify this by using our laws of exponents.
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But remember, weβre trying to find an expression for d two π¦ by dπ₯ squared.
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So weβre going to need to differentiate dπ¦ by dπ₯ with respect to π₯.
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And in this current form, we can differentiate this once again by using our general power rule.
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This time, we can see our exponent π is equal to negative one-half and our inner function π of π₯ is still equal to π₯ minus one.
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So we can use the general power rule to find an expression for our second derivative of π¦ with respect to π₯.
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This time our value of π is negative one-half, π prime is still equal to one, and π of π₯ is equal to π₯ minus one.
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And remember, since weβre multiplying our expression by one-half, we need to multiply our derivative by one-half.
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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.
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And just as we did before, we can simplify our coefficient and our exponent.
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This gives us d two π¦ by dπ₯ squared is equal to negative one-quarter times π₯ minus one all raised to the power of negative three over two.
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And we could leave our answer like this.
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However, we can also simplify this by using our laws of exponents.
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First, remember, raising a number to a negative exponent is the same as dividing by that number raised to the positive exponent.
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So, π to the power of negative three over two is equal to one divided by π to the power of three over two.
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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.
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We can use this to rewrite our answer.
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Weβll set our value of π equal to π₯ minus one.
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And this gives us our final answer.
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Therefore, we were able to show if π¦ is equal to the square root of π₯ minus one, then the second derivative of π¦ with respect to π₯ is equal to negative one divided by four times the square root of π₯ minus one all cubed.