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

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

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

tangent of

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

y = 1

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 1) : m = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

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

\frac{\partial}{\partial x} (sin (\frac{4 x}{y}))

\int x^{2} - 17 d x

\int cos^{8} (3 x) sin^{3} (3 x) d x

(3 x - 3 x^{2}) tan (x)

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