What is the tangent of y=x+sqrt(x),(49,56) ?

The tangent of y=x+sqrt(x),(49,56) is y= 15/14 x+7/2

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tangent of y=x+sqrt(x),(49,56)

tangent of

y = x + x, (49, 56)

y = \frac{15}{14} x + \frac{7}{2}

Solution steps

Find the slope of

y = x + x : \frac{d y}{d x} = 1 + \frac{1}{2 x}

Compute the slope of

y = x + x

(49, 56) : m = \frac{15}{14}

Find the line with slope m=

\frac{15}{14}

and passing through

(49, 56) : y = \frac{15}{14} x + \frac{7}{2}

y = \frac{15}{14} x + \frac{7}{2}

\int_{1}^{25} \frac{2}{x} d x

x \to + 0 lim (\frac{x ^{2}}{1 - cos ( x )})

\frac{d y}{d x} = \frac{5 x}{7 y}

\int \frac{1}{2} x cos (4 x^{2}) d x

\int - 2 x + 6 x^{2} - 4 x^{3} + C d x

What is the tangent of y=x+sqrt(x),(49,56) ?
The tangent of y=x+sqrt(x),(49,56) is y= 15/14 x+7/2