What is the tangent of f(x)=(5x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(5x)/(x^2+1),\at x=0 is y=5x

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tangent of f(x)=(5x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{5 x}{x ^{2} + 1}, at x = 0

y = 5 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{5 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{5 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

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

x = 0 : m = 5

Find the line with slope m=

5

and passing through

(0, 0) : y = 5 x

y = 5 x

\frac{d y}{d θ} = sin^{4} (π θ)

\frac{d}{d x} (e^{x} + \frac{1}{e ^{x}})

\frac{d}{d x} (∣ x - 1 ∣)

(4 y + y x^{2}) d y - (2 x + x y^{2}) d x = 0

x^{2} - x y + y^{2} = 3, (1, 2)

What is the tangent of f(x)=(5x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(5x)/(x^2+1),\at x=0 is y=5x