Question

f(x) = x$\sqrt{ x^{2} +5}$ ; (1, $\sqrt{6}$)
Find an equation of the tangent line to the graph of the function at the indicated point.

Answer 1

$\bf f(x)=x\sqrt{x^2+5}\qquad (1,\sqrt{6})\\\\ -------------------------------\\\\ \cfrac{df}{dx}=(x^2+5)^{\frac{1}{2}}+x\cdot \cfrac{1}{2}(x^2+5)^{\cfrac{}{}-\frac{1}{2}}\cdot 2x \\\\\\ \cfrac{df}{dx}=\sqrt{x^2+5}+\cfrac{x^2}{\sqrt{x^2+5}}\implies \cfrac{df}{dx}=\cfrac{x^2+5+x^2}{\sqrt{x^2+5}} \\\\\\ \left. \cfrac{df}{dx}=\cfrac{2x^2+5}{\sqrt{x^2+5}} \right|_{1,\sqrt{6}}\implies \cfrac{7}{\sqrt{6}}\implies \cfrac{7\sqrt{6}}{6}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-\sqrt{6}=\cfrac{7\sqrt{6}}{6}(x-1) \\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y=\cfrac{7\sqrt{6}}{6}x-\cfrac{7\sqrt{6}}{6}+\sqrt{6}\implies y=\cfrac{7\sqrt{6}}{6}x-\left( \cfrac{7\sqrt{6}-6\sqrt{6}}{6} \right) \\\\\\ y=\cfrac{7\sqrt{6}}{6}x-\cfrac{\sqrt{6}}{6}$