Applying Multiple Differentiation Rules

Let's consider a more complicated example.

Find the derivative of \( \displaystyle f(x) = \frac{\sqrt{x^2+1}}{x^2+3} \).

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First, notice that this is a rational function, so we'll need to use the quotient rule. The derivative of the denominator \( x^2+3 \) is simply \( 2x \). In order to use the quotient rule, however, we'll also need to know the derivative of the numerator, which we can't find directly.

In order to find the derivative of the numerator, \( \sqrt{x^2+1} \), we'll use the chain rule: \[\begin{align} \frac{d}{dx}\left( \sqrt{x^2+1} \right) &= \frac{1}{2\sqrt{x^2+1}}\cdot \frac{d}{dx}\left( x^2 + 1 \right) \\ &= \frac{1}{2\sqrt{x^2+1}}\cdot 2x \\ &= \frac{x}{\sqrt{x^2+1}} \end{align} \] Now that we know the derivatives of both the numerator and denominator, we can proceed to use the quotient rule: \[f'(x) = \frac{\left(x^2+3\right)\left( \dfrac{x}{\sqrt{x^2+1}} \right) - \sqrt{x^2+1} \ \left(2x\right)}{\left( x^2 +3 \right)^2}.\] Go through each term, and make sure you understand why it's there. You may want to refer again to the statement of the quotient rule above.

Now we should simplify, so factoring out \( \frac{1}{\sqrt{x^2+1}} \), we get:

\[f'(x) = \frac{1}{\sqrt{x^2+1}} \left(\frac{\left(x^2+3\right)\left( x \right) - \left(x^2+1\right) \left(2x\right)}{\left( x^2 +3 \right)^2}\right).\]

Multiply and combine like terms to obtain

\[ \begin{align} f'(x) &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x^3+3x-2x^3-2x}{\left( x^2 +3 \right)^2}\right) \\ &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x-x^3}{\left( x^2 +3 \right)^2}\right). \ _\square \end{align}\]