What is the tangent of y=((x+3))/((x+1)^2),\at x=0 ?

The tangent of y=((x+3))/((x+1)^2),\at x=0 is y=-5x+3

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tangent of y=((x+3))/((x+1)^2),\at x=0

tangent of

y = \frac{( x + 3 )}{( x + 1 ) ^{2}}, at x = 0

y = - 5 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

y = \frac{x + 3}{( x + 1 ) ^{2}} : \frac{d y}{d x} = \frac{- x - 5}{( x + 1 ) ^{3}}

Compute the slope of

y = \frac{x + 3}{( x + 1 ) ^{2}}

x = 0 : m = - 5

Find the line with slope m=

- 5

and passing through

(0, 3) : y = - 5 x + 3

y = - 5 x + 3

\frac{\partial}{\partial x} (3 e^{x} - 5 x e^{y})

\frac{d}{d x} (\frac{2}{1 + 3 e ^{- 2 x}})

y = sin (cos (tan (π x)))

n = 1 \sum \infty \frac{4 ^{n}}{7 ^{n}}

\frac{d}{d x} (0.006 x^{3} + 0.07 x^{2} + 18 x + 600)

What is the tangent of y=((x+3))/((x+1)^2),\at x=0 ?
The tangent of y=((x+3))/((x+1)^2),\at x=0 is y=-5x+3