compute the first two derivatives of $f(x)=sqrt{5x - 1}$. a. $f(x)=$ b. $f(x)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{5x - 1}=(5x - 1)^{\frac{1}{2}}$.

Step2: Find the first - derivative using the chain rule

The chain rule states that if $y = u^{\frac{1}{2}}$ and $u = 5x-1$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=5$. So $f'(x)=\frac{1}{2}(5x - 1)^{-\frac{1}{2}}\cdot5=\frac{5}{2\sqrt{5x - 1}}$.

Step3: Find the second - derivative using the quotient rule

The quotient rule for $y=\frac{g(x)}{h(x)}$ is $y'=\frac{g'(x)h(x)-g(x)h'(x)}{h^{2}(x)}$. Here, $g(x) = 5$ and $h(x)=2(5x - 1)^{\frac{1}{2}}$. $g'(x)=0$ and $h'(x)=2\times\frac{1}{2}(5x - 1)^{-\frac{1}{2}}\times5=\frac{5}{\sqrt{5x - 1}}$. Then $f''(x)=\frac{0\times2\sqrt{5x - 1}-5\times\frac{5}{\sqrt{5x - 1}}}{4(5x - 1)}=-\frac{25}{4(5x - 1)^{\frac{3}{2}}}$.

Answer:

a. $f'(x)=\frac{5}{2\sqrt{5x - 1}}$ b. $f''(x)=-\frac{25}{4(5x - 1)^{\frac{3}{2}}}$