# Find the second derivative of f(x)=\sqrt{2x^{3}+6x+8}.

## Question:

Find the second derivative of {eq}f(x)=\sqrt{2x^{3}+6x+8}. {/eq}

## Derivatives:

When we differentiate the function we find the rate of change of the function with respect to the variable. There are many rules of derivatives like chain rule, quotient rule, etc. Finding the second derivative is not as easy as we have to differentiate the function twice.

Chain rule of derivatives is:

{eq}\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)\\ {/eq}

Quotient rule of derivatives is

{eq}\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2} {/eq}

## Answer and Explanation:

$$f(x)=\sqrt{2x^{3}+6x+8}$$

We will differentiate the function using the chain rule of derivatives:

$$\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)\\$$

We will also use the power formula of derivatives:

$$\frac{\mathrm{d} }{\mathrm{d} x}x^{n}=nx^{n-1}\\ f'(x)=\frac{1}{2\sqrt{2x^3+6x+8}}(6x^2+6)\\ f'(x)=\frac{3(x^2+1)}{\sqrt{2x^3+6x+8}}$$

Now we will differentiate it using the quotient rule:

$$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}\\ f''(x)=\frac{d}{dx}\left(\frac{3\left(x^2+1\right)}{\sqrt{2x^3+6x+8}}\right)\\ =3\frac{d}{dx}\left(\frac{x^2+1}{\sqrt{2x^3+6x+8}}\right)\\ =3\cdot \frac{\frac{d}{dx}\left(x^2+1\right)\sqrt{2x^3+6x+8}-\frac{d}{dx}\left(\sqrt{2x^3+6x+8}\right)\left(x^2+1\right)}{\left(\sqrt{2x^3+6x+8}\right)^2}\\ =3\cdot \frac{\left(2x+1\right)\sqrt{2x^3+6x+8}-\frac{3\left(x^2+1\right)}{\sqrt{2x^3+6x+8}}\left(x^2+x\right)}{\left(\sqrt{2x^3+6x+8}\right)^2}\\ =\frac{3\left(x^3-2x^2+11x+8\right)}{2\left(x^2-x+4\right)\sqrt{2x^3+6x+8}}$$