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Calculate the differentiation
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Solve the equation
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$y = \dfrac{ a-x }{ a+x }$
$\dfrac {d } {d a } {\left( y \right)} = \dfrac { \left ( 1 - \dfrac {d } {d a } {\left( x \right)} \right ) \left ( x + a \right ) + \left ( x - a \right ) \left ( 1 + \dfrac {d } {d a } {\left( x \right)} \right ) } { x ^ { 2 } + 2 a x + a ^ { 2 } }$
Calculate the differentiation of the logarithmic function
$\dfrac {d } {d \color{#FF6800}{ a } } {\left( \color{#FF6800}{ \dfrac { a - x } { a + x } } \right)}$
$ $ Calculate the differentiation $ $
$\color{#FF6800}{ \dfrac { \left ( 1 - \dfrac {d } {d a } {\left( x \right)} \right ) \left ( x + a \right ) + \left ( x - a \right ) \left ( 1 + \dfrac {d } {d a } {\left( x \right)} \right ) } { x ^ { 2 } + 2 a x + a ^ { 2 } } }$
$\begin{cases} - x + a = x y + a y \\ x + a \neq 0 \end{cases}$
Solve the fractional equation
$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { a - x } { a + x } }$
$ $ Reverse the left and right terms of the equation (or inequality) $ $
$\color{#FF6800}{ \dfrac { a - x } { a + x } } = \color{#FF6800}{ y }$
$\color{#FF6800}{ \dfrac { a - x } { a + x } } = \color{#FF6800}{ y }$
$ $ If $ \frac{a(x)}{b(x)} = c(x) $ is valid, it is $ \begin{cases} a(x) = b(x) c(x) \\ b(x) \ne 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ x } = \left ( \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ x } \right ) \color{#FF6800}{ y } \\ \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ x } = \left ( \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ x } \right ) \color{#FF6800}{ y } \\ \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Simplify the expression $ $
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ a } = \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ y } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ a } \neq \color{#FF6800}{ 0 } \end{cases}$
Solution search results
search-thumbnail-$7.$ $\int \left(\dfrac {2a+x} {a+x}\right)\dfrac {a-x} {a+x}dx=$ $-$
10th-13th grade
Calculus
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