What is the derivative of (2^x/(1-x^2)) ?

The derivative of (2^x/(1-x^2)) is (ln(2)*2^x(1-x^2)+2^{1+x}x)/((1-x^2)^2)

What is the first derivative of (2^x/(1-x^2)) ?

The first derivative of (2^x/(1-x^2)) is (ln(2)*2^x(1-x^2)+2^{1+x}x)/((1-x^2)^2)

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

\frac{ln ( 2 ) \cdot 2 ^{x} ( 1 - x ^{2} ) + 2 ^{1 + x} x}{( 1 - x ^{2} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 ^{x} ln ( 2 ) ( 1 - x ^{2} ) - ( - 2 x ) \cdot 2 ^{x}}{( 1 - x ^{2} ) ^{2}} : \frac{ln ( 2 ) \cdot 2 ^{x} ( 1 - x ^{2} ) + 2 ^{1 + x} x}{( 1 - x ^{2} ) ^{2}}

= \frac{ln ( 2 ) \cdot 2 ^{x} ( 1 - x ^{2} ) + 2 ^{1 + x} x}{( 1 - x ^{2} ) ^{2}}

\int \frac{1}{x + !} d x

\frac{d}{d x} ((2 x - 1) (3 x + 4) (x + 7))

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

\frac{\partial}{\partial x} (+ 6 - 5 + 6 x)

n = 0 \sum \infty \frac{2}{7 ^{n}}

What is the derivative of (2^x/(1-x^2)) ?
The derivative of (2^x/(1-x^2)) is (ln(2)*2^x(1-x^2)+2^{1+x}x)/((1-x^2)^2)
What is the first derivative of (2^x/(1-x^2)) ?
The first derivative of (2^x/(1-x^2)) is (ln(2)*2^x(1-x^2)+2^{1+x}x)/((1-x^2)^2)