What is the tangent of y= 3/(sqrt(x)),(4, 3/2) ?

The tangent of y= 3/(sqrt(x)),(4, 3/2) is y=-3/16 x+9/4

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tangent of y= 3/(sqrt(x)),(4, 3/2)

tangent of

y = \frac{3}{x}, (4, \frac{3}{2})

y = - \frac{3}{16} x + \frac{9}{4}

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{3}{x}

(4, \frac{3}{2}) : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(4, \frac{3}{2}) : y = - \frac{3}{16} x + \frac{9}{4}

y = - \frac{3}{16} x + \frac{9}{4}

t^{2} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = t + e^{2 t}

f (x) = 9 x^{2}

\int - 2 x^{3} + 12 x^{2} - 20 x + 8.5 d x

\frac{d y}{d x} + 7 x e^{y} = 0

What is the tangent of y= 3/(sqrt(x)),(4, 3/2) ?
The tangent of y= 3/(sqrt(x)),(4, 3/2) is y=-3/16 x+9/4