Calculus with Applications (10th Edition)

$x=\displaystyle \pm\frac{2\sqrt{7}}{7}$
The slope of the tangent at x=a is the derivative of f(x), calculated in x=a. The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).$ \begin{align*} f(x)&=\displaystyle \frac{x}{(x^{2}+4)^{4}} \quad & \\ & \text{ quotient rule, } \left[\frac{u(x)}{v(x)}\right]^{\prime} =\displaystyle \frac{v(x)\cdot u^{\prime}(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}. & \\\\ u(x)&=x\quad u^{\prime}(x)=1 & \\ v(x)& =(x^{2}+4)^{4}\quad & \\ & \text{... apply chain rule } & \\ v^{\prime}(x)& =4(x^{2}+4)^{3}\cdot(x^{2}+4)^{\prime}\quad & \\ & =4(x^{2}+4)^{3}(2x)\quad & \\\\ f^{\prime}(x)& =\displaystyle \frac{(x^{2}+4)^{4}\cdot 1-x\cdot 4(x^{2}+4)^{3}(2x)}{(x^{2}+4)^{8}}\quad & \\ & =\displaystyle \frac{(x^{2}+4)^{3}[(x^{2}+4)-8x^{2}]}{(x^{2}+4)^{8}}\quad & \\ & =\displaystyle \frac{4-7x^{2}}{(x^{2}+4)^{5}}\quad & \\ \end{align*} The derivative is zero when \begin{align*} 4-7x^{2}&=0 \quad & \\ 4&=7x^{2} \quad & \\ x^{2}&=\displaystyle \frac{4}{7} \quad & \\ x&=\displaystyle \pm\frac{2}{\sqrt{7}}\cdot\frac{\sqrt{7}}{7}\\ x&=\displaystyle \pm\frac{2\sqrt{7}}{7} \end{align*} The tangent line is horizontal at $x=\displaystyle \pm\frac{2\sqrt{7}}{7}$