What is the tangent of y=sqrt(x),(16,4) ?

The tangent of y=sqrt(x),(16,4) is y= 1/8 x+2

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tangent of y=sqrt(x),(16,4)

tangent of

y = x, (16, 4)

y = \frac{1}{8} x + 2

Solution steps

Find the slope of

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

Compute the slope of

y = x

(16, 4) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(16, 4) : y = \frac{1}{8} x + 2

y = \frac{1}{8} x + 2

y^{^{'}} + \frac{y}{x} = \frac{7}{x ^{2}} + 3

\int (e x^{2} + π) d x

x \to 0 lim (\frac{( 2 ∣ x ∣ )}{3 x})

\int sin^{4} (x) cos (x) d x

\frac{\partial}{\partial x} (x^{2} + y^{2} - y cos (x))

What is the tangent of y=sqrt(x),(16,4) ?
The tangent of y=sqrt(x),(16,4) is y= 1/8 x+2