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

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

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

tangent of

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

Compute the slope of

f (2) = \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

t y^{^{'}} + 2 y = sin (t)

\int 6 x (2 x^{2} + 9)^{\frac{1}{2}} d x

\int \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

f^{^{'}} (x) = x^{4} + 3 x^{3} + 5 x^{2} + 2 x

\frac{d}{d x} (380000 - \frac{1}{32} x)

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