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

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

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tangent of y=sqrt(x),\at x=1

tangent of

y = x, at x = 1

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

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

y = x

x = 1 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

\int_{7}^{e^{5}} \frac{1}{x} d x

\frac{\partial}{\partial x} (sin (x \cdot y))

x \to 2 lim (2 x^{2} - 5 x - 3)

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

\frac{d}{d x} ((3 x^{2} + 8)^{5})

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