What is the tangent of y=x+sqrt(x),(1,2) ?

The tangent of y=x+sqrt(x),(1,2) is y= 3/2 x+1/2

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tangent of y=x+sqrt(x),(1,2)

tangent of

y = x + x, (1, 2)

y = \frac{3}{2} x + \frac{1}{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

(1, 2) : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 2) : y = \frac{3}{2} x + \frac{1}{2}

y = \frac{3}{2} x + \frac{1}{2}

\frac{d}{d x} (\frac{8 x ^{2} + 6 x + 4}{x})

\frac{d ^{2}}{d x ^{2}} (5 sec (x))

f (x) = 6 e^{x^{- 5}}

2 y d x + 3 x d y = 0

\int e^{2 x} 2 d x

What is the tangent of y=x+sqrt(x),(1,2) ?
The tangent of y=x+sqrt(x),(1,2) is y= 3/2 x+1/2