What is the tangent of f(x)= 8/(x^2+4),\at x=4 ?

The tangent of f(x)= 8/(x^2+4),\at x=4 is y=-4/25 x+26/25

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tangent of f(x)= 8/(x^2+4),\at x=4

tangent of

f (x) = \frac{8}{x ^{2} + 4}, at x = 4

y = - \frac{4}{25} x + \frac{26}{25}

Solution steps

Find the tangent point:

(4, \frac{2}{5})

Find the slope of

f (x) = \frac{8}{x ^{2} + 4} : \frac{df ( x )}{d x} = - \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Compute the slope of

f (x) = \frac{8}{x ^{2} + 4}

x = 4 : m = - \frac{4}{25}

Find the line with slope m=

- \frac{4}{25}

and passing through

(4, \frac{2}{5}) : y = - \frac{4}{25} x + \frac{26}{25}

y = - \frac{4}{25} x + \frac{26}{25}

\frac{\partial}{\partial y} (x^{2} + y^{2} - 4 y)

\int x arctan (16 x) d x

\frac{\partial}{\partial y} (x + \frac{x}{x ^{2} + y ^{2}})

\frac{d y}{d x} = 4 e^{- 3 x}

\int e^{x} \cdot sin (e^{x}) d x

What is the tangent of f(x)= 8/(x^2+4),\at x=4 ?
The tangent of f(x)= 8/(x^2+4),\at x=4 is y=-4/25 x+26/25