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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x + 3 e^{x}) = 1 + 3 e^{x}

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

Expand

(2 x - 2 e^{x}) (x + 3 e^{x}) - (1 + 3 e^{x}) (x^{2} - 2 e^{x}) : x^{2} - 3 e^{x} x^{2} + 4 e^{x} x + 2 e^{x}

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

\int \frac{1}{3} sec^{2} (8 x) d x

\int sin (2 t) cos (t) d t

y = (5 x - 4) (2 x^{3} - x^{2} + 1)

y^{^{''}} + 3 y^{^{'}} + 0 = 5

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

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