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If π of π₯ is equal to two π₯ cubed plus five π₯ minus two divided by π₯, find π double prime of two.
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The question wants us to find π double prime evaluated at two.
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And we recall that this is the same as finding the second derivative of π with respect to π₯ and then evaluating this at two.
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Weβll start by calculating the first derivative of our function π.
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So thatβs the derivative of two π₯ cubed plus five π₯ minus two divided by π₯ with respect to π₯.
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And we recall that for any constant π, the derivative of π multiplied by π₯ to the πth power with respect to π₯ is equal to π multiplied by π multiplied by π₯ to the power of π minus one.
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So to differentiate our first term of two π₯ cubed, we have our exponent π is equal to three.
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So differentiating this gives us two multiplied by three multiplied by π₯ to the power of three minus one.
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To differentiate our second term of five π₯, we can notice that this is actually equal to five multiplied by π₯ to the first power.
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So to differentiate this, we can apply the same rule.
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This time, we have our exponent π is equal to one.
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This gives us five multiplied by one multiplied by π₯ to the power of one minus one.
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Finally, to differentiate our third term of negative two divided by π₯, we can notice that dividing by π₯ is the same as multiplying by the reciprocal.
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This gives us a new term of negative two multiplied by π₯ to the power of negative one.
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Once again, we can apply our same rule of differentiation.
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This time, we have our exponent π is equal to negative one.
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So this gives us negative two multiplied by negative one multiplied by π₯ to the power of negative one minus one.
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Weβre now ready to simplify this expression by evaluating.
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We have two multiplied by three, which is six.
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And this is six lots of π₯ to the power of three minus one, which is two.
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And then, we add five multiplied by one, which is five lots of π₯ to the power of one minus one, which is zero.
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And then, we have negative two multiplied by negative one, which is two.
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Because a negative multiplied by a negative is a positive.
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And then, this is multiplied by π₯ to the power of negative one minus one, which is negative two.
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So weβve shown that our function π prime of π₯ is equal to six π₯ squared plus five multiplied by π₯ to the zeroth power plus two multiplied by π₯ to the power of negative two.
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And we could simplify π₯ to the zeroth power to just be equal to one.
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However, if we leave it in this form, we can differentiate it again using our differentiation rule.
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Weβre now ready to find our second derivative function.
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And we do this by finding the derivative of our first derivative function.
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So we want to find the derivative with respect to π₯ of six π₯ squared plus five π₯ to the zeroth power plus two π₯ to the power of negative two.
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And we see that we can differentiate all three of these terms by using our derivative rule.
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Our first term of six π₯ squared has an exponent π equal to two.
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So differentiating this gives us six multiplied by two multiplied by π₯ to the power of two minus one.
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Our second term of five π₯ to the zeroth power has an exponent π equal to zero.
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So differentiating this gives us five multiplied by zero multiplied by π₯ to the power of zero minus one.
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And our third term of two π₯ to the power of negative two has an exponent π equal to negative two.
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So differentiating this gives us two multiplied by negative two multiplied by π₯ to the power of negative two minus one.
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Just as we did before, we can simplify this by evaluating.
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This gives us that our second derivative function, π double prime of π₯, is equal to 12π₯ minus four π₯ to the power of negative three.
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The question wants us to find π double prime of π₯ evaluated at π₯ is equal to two.
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So we do this by substituting π₯ is equal to two into our function.
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This gives us 12 multiplied by two minus four multiplied by two to the power of negative three.
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And we recall that multiplying by a number raised to a negative power is the same as dividing by that number to the positive power.
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So four multiplied by two to the power of negative three is equal to four divided by two cubed, which is equal to four divided by eight.
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This gives us that our second derivative of π evaluated at π₯ is equal to two is equal to 24 minus four divided by eight.
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We can simplify four divided by eight to just be equal to a half.
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Instead of calculating 24 minus a half, we can change 24 into a fraction with a denominator of two.
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We do this by multiplying 24 by two to get 48 and then dividing by two.
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This gives us 48 divided by two minus a half.
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And we can evaluate this to be equal to 47 divided by two.
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So weβve shown that the second derivative of the function π of π₯ is equal to two π₯ cubed plus five π₯ minus two over π₯ at π₯ is equal to two is equal to 47 divided by two.