Mathematics: F(x) = x ; (1, ) find an equation of the tangent line to the graph of the function at the indicated point.

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F(x) = x ; (1, ) find an equation of the tangent

Ответ:

leslie4759

06.01.2022

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

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

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F(x) = x ; (1, )
find an equation of the tangent line to the graph of the function at the indicated point.

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