What is the derivative of f(x)=(x+2)/(10-3x) ?

The derivative of f(x)=(x+2)/(10-3x) is (16)/((10-3x)^2)

What is the first derivative of f(x)=(x+2)/(10-3x) ?

The first derivative of f(x)=(x+2)/(10-3x) is (16)/((10-3x)^2)

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derivative of

f (x) = \frac{x + 2}{10 - 3 x}

\frac{16}{( 10 - 3 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 2}{10 - 3 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x + 2 ) ( 10 - 3 x ) - \frac{d}{d x} ( 10 - 3 x ) ( x + 2 )}{( 10 - 3 x ) ^{2}}

\frac{d}{d x} (x + 2) = 1

\frac{d}{d x} (10 - 3 x) = - 3

= \frac{1 \cdot ( 10 - 3 x ) - ( - 3 ) ( x + 2 )}{( 10 - 3 x ) ^{2}}

Simplify

\frac{1 \cdot ( 10 - 3 x ) - ( - 3 ) ( x + 2 )}{( 10 - 3 x ) ^{2}} : \frac{16}{( 10 - 3 x ) ^{2}}

= \frac{16}{( 10 - 3 x ) ^{2}}

\int_{5}^{7} x^{3} d x

\int_{0}^{3} 3 x^{2} d x

\frac{\partial}{\partial x} (x^{\frac{2}{5}} \cdot y^{\frac{3}{5}})

\frac{d}{d x} (\frac{1}{5 + 2 x})

t an g e n t y = 5 x - 2, (5, 23)

What is the derivative of f(x)=(x+2)/(10-3x) ?
The derivative of f(x)=(x+2)/(10-3x) is (16)/((10-3x)^2)
What is the first derivative of f(x)=(x+2)/(10-3x) ?
The first derivative of f(x)=(x+2)/(10-3x) is (16)/((10-3x)^2)