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The function is $\"f(x)=\\sqrt{x}\"$ and point on the curve is $\"(x_{0},$ .

Consider $\"f(x)=\\sqrt{x}\"$ .

Apply derivative on each side with respect to $\"\\small$ .

$\"\\frac{d}{dx}f(x)=\\frac{d}{dx}(\\sqrt{x})\"$

$\"\\\\f\'(x)=\\frac{1}{2\\sqrt{x}}\"$

Slope of the tangent line at point $\"(x_{0},$ is $\"f\'(x)=\\frac{y_{0}-y}{x_{0}-x}\"$ .

Substitute $\"(x_{0},$ in the slope of the tangent line.

$\"\\\\f\'(x)=\\frac{0-y}{-4-x}$

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Slope of the tangent line is the derivative of the function.

Substitute $\"\\\\f\'(x)=\\frac{1}{2\\sqrt{x}}\"$ and $\"f(x)=\\sqrt{x}\"$ in the slope of the tangent line.

$\"\\\\\\frac{1}{2\\sqrt{x}}=\\frac{\\sqrt{x}}{4+x}$

Substitute $\"x$ in the function.

$\"\\\\f(x)=\\sqrt{4}$

The point of tangency is $\"(x,$ .

Slope of the tangent line at $\"x$ is

$\"\\\\f\'(x)=\\frac{1}{2\\sqrt{4}}$

Slope of the tangent line at $\"m=\\frac{1}{4}\"$ .

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Point slope form of line equation is $\"(y-y_{1})=m(x-x_{1})\"$ .

Substitute $\"m=\\frac{1}{4}\"$ and $\"(x,$ in the point-slope form.

$\"\\\\(y-2)=\\frac{1}{4}(x-4)$

Tangent line equation is $\"x-4y+4=0\"$ .

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The point of tangency is $\"(x,$ .

Tangent line equation is $\"x-4y+4=0\"$ .