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

The tangent of f(x)=((x^2-1))/(4x+1),\at x=1 is y= 2/5 x-2/5

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

tangent of

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

y = \frac{2}{5} x - \frac{2}{5}

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(1, 0) : y = \frac{2}{5} x - \frac{2}{5}

y = \frac{2}{5} x - \frac{2}{5}

\frac{6}{ln ( x )}

x \to 0 lim (\frac{e ^{- x} - 1}{x})

\int \frac{2 x + 3}{x ^{2} ( x + 5 ) ( x ^{2} + 1 ) ^{2}} d x

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

e^{x^{3}}

What is the tangent of f(x)=((x^2-1))/(4x+1),\at x=1 ?
The tangent of f(x)=((x^2-1))/(4x+1),\at x=1 is y= 2/5 x-2/5