What is the tangent of y= 5/(sqrt(x)),(4, 5/2) ?

The tangent of y= 5/(sqrt(x)),(4, 5/2) is y=-5/16 x+15/4

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tangent of y= 5/(sqrt(x)),(4, 5/2)

tangent of

y = \frac{5}{x}, (4, \frac{5}{2})

y = - \frac{5}{16} x + \frac{15}{4}

Solution steps

Find the slope of

y = \frac{5}{x} : \frac{d y}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{5}{x}

(4, \frac{5}{2}) : m = - \frac{5}{16}

Find the line with slope m=

- \frac{5}{16}

and passing through

(4, \frac{5}{2}) : y = - \frac{5}{16} x + \frac{15}{4}

y = - \frac{5}{16} x + \frac{15}{4}

\int \frac{e ^{x}}{e ^{x} + 1} d x

\frac{d}{d x} (0.02 e^{x} x^{3})

x (x^{2} + 4)

\frac{( x ^{2} - 1 )}{x ^{2} + 1}

f (x) = \frac{e ^{x} - e ^{x}}{e ^{x} + e ^{x}}

What is the tangent of y= 5/(sqrt(x)),(4, 5/2) ?
The tangent of y= 5/(sqrt(x)),(4, 5/2) is y=-5/16 x+15/4