66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in ...

Question

66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in the following exercises.

67. Let f(x) = √(x³ + 1).

b. Calculate f''(x).

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says let F of X be equal to the square root of the quantity of X2 + 4 in quantity. Find F of X. So we're asked to find the second derivative of F with respect to X. So, in order to do that, we first need to find our first derivative, which is going to be F of X. And we call it F of X is going to be equal to the derivative with respect to X of FX, which is the quantity of X2 + 4, in quantity raised to the 1/2. So here we've just rewritten the square root as raising that quantity to the 1/2 hour. So now we can apply our power rule and chain rule to find this derivative. So this is going to be equal to, and applying a power rule, we will have 1/2 multiplied by the quantity of X2 + 4, in quantity raised to the minus 1/2 power, but we need to apply our chain rule and multiply this by the derivative with respect to X. Of the quantity of X2 + 4 in quantity. So taking that derivative, we see that F of X is going to be equal to 1/2 multiplied by the quantity of X2 + 4 and 1D, raised to the minus 12 power. That gets multiplied by, and here when we take derivative with respect to X of X2, we get 2 X, and then we'll have plus the derivative with respect to X of 4, which is constant, so that derivative is going to be 0, so we'll have plus 0 in quantity. And just simplifying this expression, we see that FX is going to be equal to X, multiplied by the quantity of X2 + 4, in quantity raised to the minus 1/2. So now we just need to take one more derivative to find F. So F of X is going to be equal to the derivative with respect to X of F of X, which is the quantity of X multiplied by the quantity of X2 + 4 in quantity raised to the minus 1/2 in quantity. So what we need to do is apply our product rule. And here, we call that for your product rule, we have our derivative here is going to be equal to the derivative of the first terminal products, so we'll have the derivative with respect to X of X, and then that gets multiplied by the second term, which is going to be the quantity of X2 + 4 in quantity raised to the minus 12 power. Then we'll have plus the first term in our product, which was X. Multiplied by the derivative with respect to X of the second terminal product, which is the quantity of X2 + 4, in quantity raised to the minus 1/2 hour. So now we just need to take these two derivatives. So doing so, F of X is going to be equal to. When you take the derivative with respect to X of X, we get 1. And then that gets multiplied by the quantity of X2 + 4, in quantity rates to the minus 1/2 hour. Then we'll have plus X multiplied by the derivative with respect to X of the quantity of X2 + 4 and quantity raised to the minus 1/2 power. So again, here we have to apply our power rule and chain rule. So, X is going to be multiplied by the quantity of minus 1/2 multiplied by the quantity of X2 + 4, and quantity raised to the minus 3/2 power. And then that gets multiplied by the derivative with respect to X of the quantity of X2 + 4 in quantity, and we found that earlier to be to X. In quantity. So, now we just need to simplify this expression. So, the first thing we'll do is to move all of our negative exponents into the denominator. In doing so, this is going to be equal to 1, divided by the quantity of X2 + 4, and quantity raised to the 1/2 hour. Then we'll have minus. X2 divided by the quantity of X2 + 4 in quantity raised to the 3 halves power. So, now we can combine these two fractions with a common denominator. In doing so, this is going to be equal to. X2 + 4 minus X2. And that quantity is going to be divided by the quantity of X2 + 4, in quantity raised to the 3 halves power. And simplifying this expression, we see that F is going to be equal to. 4, divided by the quantity, and here we rewrite everything using a square root function. So the denominator becomes the quantity of X2 + 4. In quantity multiplied by the square root. Of the quantity of X2 + 4 in quantity. And this here is the answer that we're looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in the following exercises.
67. Let f(x) = √(x³ + 1).
b. Calculate f''(x).

Key Concepts

Second Derivative

Theorem 8.1

Function Composition and Roots

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