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

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

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tangent of y= 1/(1+x^2),(1, 1/2)

tangent of

y = \frac{1}{1 + x ^{2}}, (1, \frac{1}{2})

y = - \frac{1}{2} x + 1

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{1}{1 + x ^{2}}

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

Find the line with slope m=

- \frac{1}{2}

and passing through

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

y = - \frac{1}{2} x + 1

\frac{d}{d x} (x^{5} e^{- 3 l n (x)})

\frac{d}{d x} ((x^{2} - 5 x)^{3})

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

\frac{d y}{d x} + 5 (tan (5 x)) y = 4 cos (5 x)

x \to 0 + lim (\frac{x ^{2}}{4} - \frac{2}{x})

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