What is the derivative of y=(2x-9)/(3x+7) ?

The derivative of y=(2x-9)/(3x+7) is (41)/((3x+7)^2)

What is the first derivative of y=(2x-9)/(3x+7) ?

The first derivative of y=(2x-9)/(3x+7) is (41)/((3x+7)^2)

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derivative of

y = \frac{2 x - 9}{3 x + 7}

\frac{41}{( 3 x + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x - 9}{3 x + 7})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 x - 9 ) ( 3 x + 7 ) - \frac{d}{d x} ( 3 x + 7 ) ( 2 x - 9 )}{( 3 x + 7 ) ^{2}}

\frac{d}{d x} (2 x - 9) = 2

\frac{d}{d x} (3 x + 7) = 3

= \frac{2 ( 3 x + 7 ) - 3 ( 2 x - 9 )}{( 3 x + 7 ) ^{2}}

Expand

2 (3 x + 7) - 3 (2 x - 9) : 41

= \frac{41}{( 3 x + 7 ) ^{2}}

\int \frac{7 ^{x}}{1 - 7 ^{2 x}} d x

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

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

\frac{\partial}{\partial x} (cos (x) sin (y))

y^{^{''}} - 10 y^{^{'}} + 61 y = 0

What is the derivative of y=(2x-9)/(3x+7) ?
The derivative of y=(2x-9)/(3x+7) is (41)/((3x+7)^2)
What is the first derivative of y=(2x-9)/(3x+7) ?
The first derivative of y=(2x-9)/(3x+7) is (41)/((3x+7)^2)