What is the derivative of (x-4)/(-x^2+4) ?

The derivative of (x-4)/(-x^2+4) is (x^2-8x+4)/((-x^2+4)^2)

What is the first derivative of (x-4)/(-x^2+4) ?

The first derivative of (x-4)/(-x^2+4) is (x^2-8x+4)/((-x^2+4)^2)

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derivative of

\frac{x - 4}{- x ^{2} + 4}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

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

\frac{\partial}{\partial x} (\frac{60}{1 + x ^{2} + y ^{2}})

\frac{d y}{d x} = 5 + y - 5 x + 2

x^{2} y^{^{'}} + \frac{4}{3} x y = 2 cos (3 x) y^{- \frac{1}{2}}

What is the derivative of (x-4)/(-x^2+4) ?
The derivative of (x-4)/(-x^2+4) is (x^2-8x+4)/((-x^2+4)^2)
What is the first derivative of (x-4)/(-x^2+4) ?
The first derivative of (x-4)/(-x^2+4) is (x^2-8x+4)/((-x^2+4)^2)