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

The derivative of-(3y/(3x+2y)) is -(3(3x\frac{dy)/(dx)-3y)}{(3x+2y)^2}

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

The first derivative of-(3y/(3x+2y)) is -(3(3x\frac{dy)/(dx)-3y)}{(3x+2y)^2}

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x + 2 y) = 3 + 2 \frac{d y}{d x}

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

Simplify

- 3 \cdot \frac{\frac{d y}{d x} ( 3 x + 2 y ) - ( 3 + 2 \frac{d y}{d x} ) y}{( 3 x + 2 y ) ^{2}} : - \frac{3 ( 3 x \frac{d y}{d x} - 3 y )}{( 3 x + 2 y ) ^{2}}

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

\frac{d}{d x} (\frac{1}{( 3 - x ) ^{2}})

cos (k \cdot t)

\frac{d y}{d x} = y^{\frac{1}{3}}, y (1) = 0

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

f (3) = 10 + 12 x - 3 x^{2} - x^{3}

What is the derivative of-(3y/(3x+2y)) ?
The derivative of-(3y/(3x+2y)) is -(3(3x\frac{dy)/(dx)-3y)}{(3x+2y)^2}
What is the first derivative of-(3y/(3x+2y)) ?
The first derivative of-(3y/(3x+2y)) is -(3(3x\frac{dy)/(dx)-3y)}{(3x+2y)^2}