Question

F(x) = x
`\sqrt{ x^(2) +5}` ; (1,
`√(6)`)

Find an equation of the tangent line to the graph of the function at the indicated point.

Answer 1

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


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