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find the equation of the tangent line to the curve when x has the given value. f(x) = sqrt (x) ; x= 49
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Can you find the derivative of y = x^(1/2)
The line will be y - f(49) = f ' (49)*(x - 49)
the slope of the tangent line is the first derivative of the curve at that point
\[\frac{ d }{ dx }\sqrt{x} = \frac{ d }{ dx }x ^{1/2} = \frac{ 1 }{ 2 }x ^{1/2 - 1} = \frac{ 1 }{ 2 }x ^{-1/2} = \frac{ 1 }{ 2\sqrt{x} }\]
Just used the power rule for derivatives. \[\frac{ d }{ dx }x^n = n*x ^{n-1}\]
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With the point x=49, y = f(49) = sqrt(49) = 7 point; (49,7) and slope m = f '(49) = 1/(2*7) = 1/14 Form Line from Point-Slope Form y - f(49) = f '(49) *(x-49)
u understand everything... @Ko-Ki
i think so thx
here is the graph
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thanks
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