Find an equation of the tangent line to the curve at the given point.
$y = \sqrt{x}, (81, 9)$

Question

Find an equation of the tangent line to the curve at the given point.  y=x, (81,9)

reinosodairyshm · Accepted Answer

y=x, (81,9)  y=x  Differentiating with respect x, we get  dydx=12x  dydx=1281=12(9)=118  The equation of tangent with slope 118 and the point (81,9) is  y−9=118(x−81)  y−9=118x−8118  y−9=118x−92  y=118x−92+9  y=118x+−9+182  y=118x+92

RizerMix · Accepted Answer

Find dydx and evaluate at x=81 and y=9 to find the slope of the tangent line at x=81 and y=9Use axn=axn to rewrite x as x12y=x12Differentiate both sides of the equation.ddx(y)=ddx(x12)The derivative of y with respect to x is y′y′Differentiate the right side of the equation.12x12Reform the equation by setting the left side equal to the right side.y′=12x12Replace y&#039; with dydxdydx=12x12Evaluate at x=81 and y=9.118Plug in the slope of the tangent line and the x and y values of the point into the point-slope formula y−y1=n(x−x1)y−(9)=118⋅(x−(81))Simplify,y=118x+92

Jeffrey Jordon · Accepted Answer

Great two answers. Thank you.

Find an equation of the tangent line to the curve

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