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

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

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

tangent of

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

y = - 3 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Find the line with slope m=

- 3

and passing through

(0, 1) : y = - 3 x + 1

y = - 3 x + 1

\frac{\partial}{\partial y} (A x + B y^{2} + C x^{2} y)

y d x + (y - x) d y = 0

(0.3) (2.5)

\int x^{5} \cdot ln (3 x^{2}) d x

\int \frac{1}{x + 5} d x

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