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

The tangent of y=sqrt(x),\at x=4 is y= 1/4 x+1

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tangent of y=sqrt(x),\at x=4

tangent of

y = x, at x = 4

y = \frac{1}{4} x + 1

Solution steps

Find the tangent point:

(4, 2)

Find the slope of

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

Compute the slope of

y = x

x = 4 : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

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

y = \frac{1}{4} x + 1

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

y^{^{'''}} + 4 y^{^{''}} + 4 y^{^{'}} = 8 x

\int x^{4} e^{\frac{x}{2}} d x

\frac{\partial}{\partial x} (2 g (x) x)

\frac{d y}{d x} = \frac{x}{9 y}

What is the tangent of y=sqrt(x),\at x=4 ?
The tangent of y=sqrt(x),\at x=4 is y= 1/4 x+1