What is the derivative of (3x^2e^x-4/(x^2)) ?

The derivative of (3x^2e^x-4/(x^2)) is (3e^xx^3+8)/(x^3)

What is the first derivative of (3x^2e^x-4/(x^2)) ?

The first derivative of (3x^2e^x-4/(x^2)) is (3e^xx^3+8)/(x^3)

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

\frac{3 e ^{x} x ^{3} + 8}{x ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{3 e ^{x} x ^{3} + 8}{x ^{3}}

5 y^{^{'}} = y^{2}, y (0) = 1

\int_{0}^{1} \frac{x ^{2}}{x ^{2} + 1} d x

\frac{d}{d x} (1 - \frac{2}{x})

y^{^{''}} + x y^{^{'}} - y = 0

\frac{d}{d x} (- \frac{cos ( x )}{x})

What is the derivative of (3x^2e^x-4/(x^2)) ?
The derivative of (3x^2e^x-4/(x^2)) is (3e^xx^3+8)/(x^3)
What is the first derivative of (3x^2e^x-4/(x^2)) ?
The first derivative of (3x^2e^x-4/(x^2)) is (3e^xx^3+8)/(x^3)