What is the tangent of y=x^3,\at x=-1 ?

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

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tangent of y=x^3,\at x=-1

tangent of

y = x^{3}, at x = - 1

y = 3 x + 2

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

y = x^{3} : \frac{d y}{d x} = 3 x^{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

(- 1, - 1) : y = 3 x + 2

y = 3 x + 2

f (x) = \frac{2 x}{sin ( x ) + cos ( x )}

\int \frac{( y ^{2} + 1 ) ^{2}}{y ^{2}} d y

x \to 2 lim (\frac{3}{x})

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

\int 3 x^{2} y + e^{x} sin (y) d x