What is the tangent of f(x)= 1/(x^2+4),(1, 1/5) ?

The tangent of f(x)= 1/(x^2+4),(1, 1/5) is y=-2/25 x+7/25

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tangent of f(x)= 1/(x^2+4),(1, 1/5)

tangent of

f (x) = \frac{1}{x ^{2} + 4}, (1, \frac{1}{5})

y = - \frac{2}{25} x + \frac{7}{25}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{2} + 4} : \frac{df ( x )}{d x} = - \frac{2 x}{( x ^{2} + 4 ) ^{2}}

Compute the slope of

f (x) = \frac{1}{x ^{2} + 4}

(1, \frac{1}{5}) : m = - \frac{2}{25}

Find the line with slope m=

- \frac{2}{25}

and passing through

(1, \frac{1}{5}) : y = - \frac{2}{25} x + \frac{7}{25}

y = - \frac{2}{25} x + \frac{7}{25}

\int 3 e^{x} - 3 x e^{x^{2}} d x

\frac{d}{d x} (k x + d)

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

\int \frac{1}{2 x ^{3} - 3 x ^{2}} d x

What is the tangent of f(x)= 1/(x^2+4),(1, 1/5) ?
The tangent of f(x)= 1/(x^2+4),(1, 1/5) is y=-2/25 x+7/25