Question

Evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. (Round your answer to three decimal places.)f(x) = 3x + 9 (0, 1)f ''(0) =

Answer 1

We want to evaluate the second derivative of the following function

$f(x)=\frac{3}{\sqrt{x+9}}$

at x = 0.

To calculate the derivative of this function, we just have to use the power rule and along with the chain rule. Those rules are

$\begin{gathered} \frac{d}{dx}x^n=nx^{n-1} \\ (f(g(x)))^{\prime}=f^{\prime}(g(x))g^{\prime}(x) \end{gathered}$

Then, we have

$\begin{gathered} f(x)=\frac{3}{\sqrt{x+9}}=3(x+9)^{-1/2} \\ f^{\prime}(x)=3\cdot(-\frac{1}{2})(x+9)^{-1/2-1}\cdot(x+9)^{\prime}=\frac{3}{2}(x+9)^{-3/2} \\ f^{\prime\prime}(x)=-\frac{3}{2}\cdot(-\frac{3}{2})(x+9)^{-3/2-1}\cdot(x+9)^{\prime}=-\frac{9}{4}(x+9)^{-5/2} \\ f^{^{\prime\prime^{\prime}}}(x)=\frac{9}{4}\frac{1}{\sqrt{(x+9)^5}} \end{gathered}$

Now that we have the function, we just have to evaluate the function at x = 0.

$f^{\prime\prime}(0)=\frac{9}{4}\frac{1}{\sqrt{(0+9)^5}}=\frac{9}{4}\frac{1}{9^2\sqrt{9}}=\frac{9}{4\cdot9^2\cdot3}=\frac{1}{108}\approx0.009$

TThe second derivative of this function evaluated at x = 0 is 0.009.