What is the ((e^{4x^2})/(x^2))^' ?

The ((e^{4x^2})/(x^2))^' is (2e^{4x^2}(4x^2-1))/(x^3)

What is the first ((e^{4x^2})/(x^2))^' ?

The first ((e^{4x^2})/(x^2))^' is (2e^{4x^2}(4x^2-1))/(x^3)

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(\frac{e ^{4 x^{2}}}{x ^{2}})^{^{'}}

\frac{2 e ^{4 x^{2}} ( 4 x ^{2} - 1 )}{x ^{3}}

Solution steps

(\frac{e ^{4 x^{2}}}{x ^{2}})^{'}

Apply the Quotient Rule:

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

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

(e^{4 x^{2}})^{'} = e^{4 x^{2}} \cdot 8 x

(x^{2})^{'} = 2 x

= \frac{e ^{4 x^{2}} \cdot 8 x x ^{2} - 2 x e ^{4 x^{2}}}{( x ^{2} ) ^{2}}

Simplify

\frac{e ^{4 x^{2}} \cdot 8 x x ^{2} - 2 x e ^{4 x^{2}}}{( x ^{2} ) ^{2}} : \frac{2 e ^{4 x^{2}} ( 4 x ^{2} - 1 )}{x ^{3}}

= \frac{2 e ^{4 x^{2}} ( 4 x ^{2} - 1 )}{x ^{3}}

\int \frac{( 7 x ^{9} - 20 x ^{6} )}{x ^{3}} d x

\int (tan (x) + \frac{1}{tan ( x )})^{2} d x

x^{2} y^{^{''}} - 2 y = sin (ln (x))

x \to - 2 lim (2 x)

\int \frac{1}{5 x ^{7}} d x

What is the ((e^{4x^2})/(x^2))^' ?
The ((e^{4x^2})/(x^2))^' is (2e^{4x^2}(4x^2-1))/(x^3)
What is the first ((e^{4x^2})/(x^2))^' ?
The first ((e^{4x^2})/(x^2))^' is (2e^{4x^2}(4x^2-1))/(x^3)