How do you find the first and second derivative of $y = e^{x^{3}}$ $y=e^(x^3)$ ?

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How do you find the first and second derivative of $y = e^{x^{3}}$ $y=e^(x^3)$ ?

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Answer 1 · 2016-12-15T17:24:49

$\frac{d}{d x} e^{x^{3}} = 3 x^{2} e^{x^{3}}$ $d/(dx) e^(x^3)= 3x^2e^(x^3)$
$\frac{d^{(2)}}{{d x}^{2}} e^{x^{3}} = 6 x e^{x^{3}} (1 + \frac{3}{2} x^{3})$ $d^((2))/(dx^2) e^(x^3)=6xe^(x^3)(1+3/2x^3)$

Explanation:

Using the chain rule:

$\frac{d}{d x} e^{x^{3}} = \frac{d e^{x^{3}}}{d (x^{3})} \cdot \frac{d (x^{3})}{d x} = 3 x^{2} e^{x^{3}}$ $d/(dx) e^(x^3) = (de^(x^3))/(d(x^3)) * (d(x^3)) /(dx) = 3x^2e^(x^3)$

Then using the product rule:

$\frac{d^{(2)}}{{d x}^{2}} e^{x^{3}} = \frac{d}{d x} (3 x^{2} e^{x^{3}}) = (\frac{d}{d x} 3 x^{2}) e^{x^{3}} + 3 x^{2} (\frac{d}{d x} e^{x^{3}}) = 6 x e^{x^{3}} + 3 x^{2} \cdot 3 x^{2} e^{x^{3}} = 6 x e^{x^{3}} + 9 x^{4} e^{x^{3}} = 6 x e^{x^{3}} (1 + \frac{3}{2} x^{3})$ $d^((2))/(dx^2) e^(x^3) =d/(dx)(3x^2e^(x^3)) = (d/(dx)3x^2)e^(x^3)+ 3x^2(d/(dx) e^(x^3)) = 6xe^(x^3)+3x^2*3x^2e^(x^3)=6xe^(x^3)+9x^4e^(x^3)=6xe^(x^3)(1+3/2x^3)$

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How do you find the first and second derivative of $y = e^{x^{3}}$ $y=e^(x^3)$ ?

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How do you find the first and second derivative of $y = e^{x^{3}}$ $y=e^(x^3)$ ?