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

The tangent of y= 1/(sqrt(x)),(4, 1/2) is y=-1/16 x+3/4

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tangent of y= 1/(sqrt(x)),(4, 1/2)

tangent of

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

y = - \frac{1}{16} x + \frac{3}{4}

Solution steps

Find the slope of

y = \frac{1}{x} : \frac{d y}{d x} = - \frac{1}{2 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{x}

(4, \frac{1}{2}) : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(4, \frac{1}{2}) : y = - \frac{1}{16} x + \frac{3}{4}

y = - \frac{1}{16} x + \frac{3}{4}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 t}

\frac{\partial}{\partial x} (x^{s i n (x)})

6 \cdot (\frac{d y}{d x}) = y^{2}, y (0) = 4

y = x^{3} - 2 x + 2, (3, 23)

\frac{d y}{d x} = \frac{( 3 y ^{2} )}{x}

What is the tangent of y= 1/(sqrt(x)),(4, 1/2) ?
The tangent of y= 1/(sqrt(x)),(4, 1/2) is y=-1/16 x+3/4