What is the derivative of (Ax+B/(sqrt(4-x))) ?

The derivative of (Ax+B/(sqrt(4-x))) is (8A-Ax+B)/(2(4-x)sqrt(4-x))

What is the first derivative of (Ax+B/(sqrt(4-x))) ?

The first derivative of (Ax+B/(sqrt(4-x))) is (8A-Ax+B)/(2(4-x)sqrt(4-x))

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\frac{d}{d x} (\frac{A x + B}{4 - x})

\frac{8 A - A x + B}{2 ( 4 - x ) 4 - x}

Solution steps

\frac{d}{d x} (\frac{A x + B}{4 - x})

Treat

A, B

as constants

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( A x + B ) 4 - x - \frac{d}{d x} ( 4 - x ) ( A x + B )}{( 4 - x ) ^{2}}

\frac{d}{d x} (A x + B) = A

\frac{d}{d x} (4 - x) = - \frac{1}{2 4 - x}

= \frac{A 4 - x - ( - \frac{1}{2 4 - x} ) ( A x + B )}{( 4 - x ) ^{2}}

Simplify

\frac{A 4 - x - ( - \frac{1}{2 4 - x} ) ( A x + B )}{( 4 - x ) ^{2}} : \frac{8 A - A x + B}{2 ( 4 - x ) 4 - x}

= \frac{8 A - A x + B}{2 ( 4 - x ) 4 - x}

n = 1 \sum \infty \frac{3}{n !}

\frac{\partial}{\partial x} (4 x y^{2})

x \to 3 lim (9 - 9)

\frac{\partial}{\partial x} (\frac{x y}{x ^{2} - y ^{2}})

\frac{\partial}{\partial x} (\frac{a y x ^{2} z}{2})

What is the derivative of (Ax+B/(sqrt(4-x))) ?
The derivative of (Ax+B/(sqrt(4-x))) is (8A-Ax+B)/(2(4-x)sqrt(4-x))
What is the first derivative of (Ax+B/(sqrt(4-x))) ?
The first derivative of (Ax+B/(sqrt(4-x))) is (8A-Ax+B)/(2(4-x)sqrt(4-x))