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

The tangent of y=e^{x^3},(1,e) is y=3ex-2e

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

tangent of

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

y = 3 e x - 2 e

Solution steps

Find the slope of

y = e^{x^{3}} : \frac{d y}{d x} = e^{x^{3}} \cdot 3 x^{2}

Compute the slope of

y = e^{x^{3}}

(1, e) : m = 3 e

Find the line with slope m=

3 e

and passing through

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

y = 3 e x - 2 e

\frac{e ^{x}}{6 + e ^{x}}

\frac{d}{d x} (\frac{1}{3 x - 4})

\frac{d}{d x} (2.2716 e^{0.2865 x})

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

\int_{0}^{\infty} sin (x) d x

What is the tangent of y=e^{x^3},(1,e) ?
The tangent of y=e^{x^3},(1,e) is y=3ex-2e