Video: Find the Second Derivative of the Sum of Rational Functions

If 𝑓(π‘₯) = 2π‘₯Β³ + 5π‘₯ βˆ’ (2/π‘₯), find 𝑓″(2).

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Video Transcript

If 𝑓 of π‘₯ is equal to two π‘₯ cubed plus five π‘₯ minus two divided by π‘₯, find 𝑓 double prime of two.

The question wants us to find 𝑓 double prime evaluated at two. And we recall that this is the same as finding the second derivative of 𝑓 with respect to π‘₯ and then evaluating this at two. We’ll start by calculating the first derivative of our function 𝑓. So that’s the derivative of two π‘₯ cubed plus five π‘₯ minus two divided by π‘₯ with respect to π‘₯. And we recall that for any constant π‘Ž, the derivative of π‘Ž multiplied by π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž multiplied by 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one.

So to differentiate our first term of two π‘₯ cubed, we have our exponent 𝑛 is equal to three. So differentiating this gives us two multiplied by three multiplied by π‘₯ to the power of three minus one. 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. So to differentiate this, we can apply the same rule. This time, we have our exponent 𝑛 is equal to one. This gives us five multiplied by one multiplied by π‘₯ to the power of one minus one. 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. This gives us a new term of negative two multiplied by π‘₯ to the power of negative one.

Once again, we can apply our same rule of differentiation. This time, we have our exponent 𝑛 is equal to negative one. So this gives us negative two multiplied by negative one multiplied by π‘₯ to the power of negative one minus one. We’re now ready to simplify this expression by evaluating. We have two multiplied by three, which is six. And this is six lots of π‘₯ to the power of three minus one, which is two. And then, we add five multiplied by one, which is five lots of π‘₯ to the power of one minus one, which is zero. And then, we have negative two multiplied by negative one, which is two. Because a negative multiplied by a negative is a positive. And then, this is multiplied by π‘₯ to the power of negative one minus one, which is negative two.

So we’ve shown that our function 𝑓 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. And we could simplify π‘₯ to the zeroth power to just be equal to one. However, if we leave it in this form, we can differentiate it again using our differentiation rule. We’re now ready to find our second derivative function. And we do this by finding the derivative of our first derivative function.

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. And we see that we can differentiate all three of these terms by using our derivative rule. Our first term of six π‘₯ squared has an exponent 𝑛 equal to two. So differentiating this gives us six multiplied by two multiplied by π‘₯ to the power of two minus one. Our second term of five π‘₯ to the zeroth power has an exponent 𝑛 equal to zero.

So differentiating this gives us five multiplied by zero multiplied by π‘₯ to the power of zero minus one. And our third term of two π‘₯ to the power of negative two has an exponent 𝑛 equal to negative two. So differentiating this gives us two multiplied by negative two multiplied by π‘₯ to the power of negative two minus one. Just as we did before, we can simplify this by evaluating. This gives us that our second derivative function, 𝑓 double prime of π‘₯, is equal to 12π‘₯ minus four π‘₯ to the power of negative three.

The question wants us to find 𝑓 double prime of π‘₯ evaluated at π‘₯ is equal to two. So we do this by substituting π‘₯ is equal to two into our function. This gives us 12 multiplied by two minus four multiplied by two to the power of negative three. 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. 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. This gives us that our second derivative of 𝑓 evaluated at π‘₯ is equal to two is equal to 24 minus four divided by eight.

We can simplify four divided by eight to just be equal to a half. Instead of calculating 24 minus a half, we can change 24 into a fraction with a denominator of two. We do this by multiplying 24 by two to get 48 and then dividing by two. This gives us 48 divided by two minus a half. And we can evaluate this to be equal to 47 divided by two. So we’ve shown that the second derivative of the function 𝑓 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.

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