`y = sqrt(1 + x^3), (2, 3)` Find an equation of the tangent line to the curve at the given point.
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gsarora17
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Slope of a tangent line to a function at a point is equal to the derivative of the function at that point.
`y=(1+x^3)^(1/2)`
`dy/dx=(1/2)*(1+x^3)^((1/2)-1)*3x^2`
`dy/dx=(3x^2)/(2(sqrt(1+x^3)))`
Slope (m) at (2,3) = (3*2^2)/(2*3) = 2` `
Equation of tangent line can be found by using point slope form of the line.
y-y_1=m(x-x_1)
y-3=2(x-2)
y-3=2x-4
y=2x-1
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