What is the derivative of ((4-y))/((2^y+3)) ?

The derivative of ((4-y))/((2^y+3)) is (-2^y-3-ln(2)*2^{2+y}+ln(2)*2^yy)/((2^y+3)^2)

What is the first derivative of ((4-y))/((2^y+3)) ?

The first derivative of ((4-y))/((2^y+3)) is (-2^y-3-ln(2)*2^{2+y}+ln(2)*2^yy)/((2^y+3)^2)

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derivative of

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

\frac{- 2 ^{y} - 3 - ln ( 2 ) \cdot 2 ^{2 + y} + ln ( 2 ) \cdot 2 ^{y} y}{( 2 ^{y} + 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d y} (4 - y) = - 1

\frac{d}{d y} (2^{y} + 3) = ln (2) \cdot 2^{y}

= \frac{( - 1 ) ( 2 ^{y} + 3 ) - ln ( 2 ) \cdot 2 ^{y} ( 4 - y )}{( 2 ^{y} + 3 ) ^{2}}

Simplify

\frac{( - 1 ) ( 2 ^{y} + 3 ) - ln ( 2 ) \cdot 2 ^{y} ( 4 - y )}{( 2 ^{y} + 3 ) ^{2}} : \frac{- 2 ^{y} - 3 - ln ( 2 ) \cdot 2 ^{2 + y} + ln ( 2 ) \cdot 2 ^{y} y}{( 2 ^{y} + 3 ) ^{2}}

= \frac{- 2 ^{y} - 3 - ln ( 2 ) \cdot 2 ^{2 + y} + ln ( 2 ) \cdot 2 ^{y} y}{( 2 ^{y} + 3 ) ^{2}}

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

\int sin^{3} (3 x) cos^{- 2} (3 x) d x

\frac{d}{d x} ((\frac{x}{7} + \frac{7}{x})^{7})

\int \frac{1}{3 x ^{2} - 1} d x

i = 2 \sum \infty \frac{5}{i ^{2} + i - 2}

What is the derivative of ((4-y))/((2^y+3)) ?
The derivative of ((4-y))/((2^y+3)) is (-2^y-3-ln(2)*2^{2+y}+ln(2)*2^yy)/((2^y+3)^2)
What is the first derivative of ((4-y))/((2^y+3)) ?
The first derivative of ((4-y))/((2^y+3)) is (-2^y-3-ln(2)*2^{2+y}+ln(2)*2^yy)/((2^y+3)^2)