What is the derivative of f(x)=(e^{x^2})/(x^3) ?

The derivative of f(x)=(e^{x^2})/(x^3) is (e^{x^2}(2x^2-3))/(x^4)

What is the first derivative of f(x)=(e^{x^2})/(x^3) ?

The first derivative of f(x)=(e^{x^2})/(x^3) is (e^{x^2}(2x^2-3))/(x^4)

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derivative of

f (x) = \frac{e ^{x^{2}}}{x ^{3}}

\frac{e ^{x^{2}} ( 2 x ^{2} - 3 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{e ^{x^{2}}}{x ^{3}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( e ^{x^{2}} ) x ^{3} - \frac{d}{d x} ( x ^{3} ) e ^{x^{2}}}{( x ^{3} ) ^{2}}

\frac{d}{d x} (e^{x^{2}}) = e^{x^{2}} \cdot 2 x

\frac{d}{d x} (x^{3}) = 3 x^{2}

= \frac{e ^{x^{2}} \cdot 2 x x ^{3} - 3 x ^{2} e ^{x^{2}}}{( x ^{3} ) ^{2}}

Simplify

\frac{e ^{x^{2}} \cdot 2 x x ^{3} - 3 x ^{2} e ^{x^{2}}}{( x ^{3} ) ^{2}} : \frac{e ^{x^{2}} ( 2 x ^{2} - 3 )}{x ^{4}}

= \frac{e ^{x^{2}} ( 2 x ^{2} - 3 )}{x ^{4}}

x \to 0 lim (x ln (1 - \frac{2}{3 w ( x )}))

x \to - 3 lim (2 x^{2} - 3 x + 4)

\frac{d}{d x} (x^{3} - 9 x^{2} + 24 x - 15)

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x), y (0) = 0, y^{^{'}} (0) = 0

\int \frac{u}{1 + u ^{2}} d u

What is the derivative of f(x)=(e^{x^2})/(x^3) ?
The derivative of f(x)=(e^{x^2})/(x^3) is (e^{x^2}(2x^2-3))/(x^4)
What is the first derivative of f(x)=(e^{x^2})/(x^3) ?
The first derivative of f(x)=(e^{x^2})/(x^3) is (e^{x^2}(2x^2-3))/(x^4)