What is the tangent of g(x)=sqrt(x),(1,1) ?

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

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tangent of g(x)=sqrt(x),(1,1)

tangent of

g (x) = x, (1, 1)

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

Solution steps

Find the slope of

g (x) = x : \frac{d g ( x )}{d x} = \frac{1}{2 x}

Compute the slope of

g (x) = x

(1, 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}

\frac{\partial}{\partial x} (3 x cos (4 x y))

\frac{\partial}{\partial x} (9 x e^{8 x y})

\int x (x + 1)^{8} d x

\frac{d y}{d x} + y = x e^{5 x} y^{6}

\int_{- 2}^{2} (x + 5) d x

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