What is the tangent of y=e^{x^3},(-1, 1/e) ?

The tangent of y=e^{x^3},(-1, 1/e) is y= 3/e x+4/e

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tangent of y=e^{x^3},(-1, 1/e)

tangent of

y = e^{x^{3}}, (- 1, \frac{1}{e})

y = \frac{3}{e} x + \frac{4}{e}

Solution steps

Find the slope of

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

Find the line with slope m=

\frac{3}{e}

and passing through

(- 1, \frac{1}{e}) : y = \frac{3}{e} x + \frac{4}{e}

y = \frac{3}{e} x + \frac{4}{e}

\frac{( y - 2 ) ^{3}}{( y ^{2} + 4 y ) ^{7}}

x \to + 3 lim (f (x))

x \to 90 lim (sin (x))

\frac{d y}{d x} = \frac{x}{16 y}

\frac{d y}{d x} = 7 x y

What is the tangent of y=e^{x^3},(-1, 1/e) ?
The tangent of y=e^{x^3},(-1, 1/e) is y= 3/e x+4/e