What is the derivative of y=(e^x)/(9x^2) ?

The derivative of y=(e^x)/(9x^2) is (e^x(x-2))/(9x^3)

What is the first derivative of y=(e^x)/(9x^2) ?

The first derivative of y=(e^x)/(9x^2) is (e^x(x-2))/(9x^3)

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derivative of

y = \frac{e ^{x}}{9 x ^{2}}

\frac{e ^{x} ( x - 2 )}{9 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{9 x ^{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{9} \frac{d}{d x} (\frac{e ^{x}}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{1}{9} \cdot \frac{\frac{d}{d x} ( e ^{x} ) x ^{2} - \frac{d}{d x} ( x ^{2} ) e ^{x}}{( x ^{2} ) ^{2}}

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

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

= \frac{1}{9} \cdot \frac{e ^{x} x ^{2} - 2 x e ^{x}}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{9} \cdot \frac{e ^{x} x ^{2} - 2 x e ^{x}}{( x ^{2} ) ^{2}} : \frac{e ^{x} ( x - 2 )}{9 x ^{3}}

= \frac{e ^{x} ( x - 2 )}{9 x ^{3}}

\int \frac{( x ^{2} )}{13 - 4 x + x ^{2}} d x

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 1})

y^{^{''}} + 4 y^{^{'}} + 3 y = t e^{- t}

x \to 6 lim (\frac{3 x}{x - 6})

\int r^{2} e^{r} d r

What is the derivative of y=(e^x)/(9x^2) ?
The derivative of y=(e^x)/(9x^2) is (e^x(x-2))/(9x^3)
What is the first derivative of y=(e^x)/(9x^2) ?
The first derivative of y=(e^x)/(9x^2) is (e^x(x-2))/(9x^3)