What is the tangent of 5/(sqrt(x)),\at x= 1/4 ?

The tangent of 5/(sqrt(x)),\at x= 1/4 is y=-20x+15

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tangent of 5/(sqrt(x)),\at x= 1/4

tangent of

\frac{5}{x}, at x = \frac{1}{4}

y = - 20 x + 15

Solution steps

Find the tangent point:

(\frac{1}{4}, 10)

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}

x = \frac{1}{4} : m = - 20

Find the line with slope m=

- 20

and passing through

(\frac{1}{4}, 10) : y = - 20 x + 15

y = - 20 x + 15

cos (x - 1)

\frac{d}{d t} (sin^{2} (e^{s i n^{2} (t)}))

\frac{\partial}{\partial z} (\frac{z ^{2}}{y})

\frac{d}{d x} (x^{2} - cos (2 x))

(12 + x^{2}) y^{^{'}} - 2 x y = 0

What is the tangent of 5/(sqrt(x)),\at x= 1/4 ?
The tangent of 5/(sqrt(x)),\at x= 1/4 is y=-20x+15