Product rule and chain rule together

The expression under the square root is always positive, $x^2 + 9 \geq 9 > 0$, so the function is defined and differentiable for every real number, $D = \mathbb{R}$. To use the product rule, write it as a product of two factors:

The first factor is just $x$, and the derivative of $x$ is $1$:

The factor $v$ is a function inside a function, so it needs the chain rule. The outer function is the square root, whose derivative is $\frac{1}{2\sqrt{t}}$. The inner function is $t = x^2 + 9$, whose derivative is $2x$. Multiply the derivative of the outer function (with the inside left untouched) by the derivative of the inner function:

Cancel the $2$:

Now put $u$, $u'$, $v$ and $v'$ into the product rule, $(u \cdot v)' = u' \cdot v + u \cdot v'$:

To add the two terms, rewrite the first one as a fraction with denominator $\sqrt{x^2 + 9}$. This works because $\sqrt{x^2 + 9} \cdot \sqrt{x^2 + 9} = x^2 + 9$:

Now add the numerators:

Substitute $x = 4$ into the simplified derivative. Start with the square root in the denominator:

Then the numerator:

So: