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

The tangent of f(x)= 4/(1+x^2),\at x=1 is y=-2x+4

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

tangent of

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

y = - 2 x + 4

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = \frac{4}{1 + x ^{2}} : \frac{df ( x )}{d x} = - \frac{8 x}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

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

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 2) : y = - 2 x + 4

y = - 2 x + 4

x \to \frac{π}{2} - lim (e^{s e c (x)})

\int \frac{e t}{t} d t

\int \frac{x ^{2}}{4 + x ^{2}} d x

\frac{\partial}{\partial x} (y cos (π x^{2} y))

f (x) = cos (e^{x})

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