What is the derivative of (x^2-4/(x+0.5)) ?

The derivative of (x^2-4/(x+0.5)) is (x^2+x+4)/((x+0.5)^2)

What is the first derivative of (x^2-4/(x+0.5)) ?

The first derivative of (x^2-4/(x+0.5)) is (x^2+x+4)/((x+0.5)^2)

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\frac{d}{d x} (\frac{x ^{2} - 4}{x + 0.5})

\frac{x ^{2} + x + 4}{( x + 0.5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 4}{x + 0.5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 4 ) ( x + 0.5 ) - \frac{d}{d x} ( x + 0.5 ) ( x ^{2} - 4 )}{( x + 0.5 ) ^{2}}

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

\frac{d}{d x} (x + 0.5) = 1

= \frac{2 x ( x + 0.5 ) - 1 \cdot ( x ^{2} - 4 )}{( x + 0.5 ) ^{2}}

Simplify

\frac{2 x ( x + 0.5 ) - 1 \cdot ( x ^{2} - 4 )}{( x + 0.5 ) ^{2}} : \frac{x ^{2} + x + 4}{( x + 0.5 ) ^{2}}

= \frac{x ^{2} + x + 4}{( x + 0.5 ) ^{2}}

y \to 0 lim (a x^{2} + b x^{3})

f (x) = 2 e^{x} (cos (x) - sin (x))

30 y^{^{''}} + 19 y^{^{'}} - 5 y = 0

(3, 4), (5, - 4)

\frac{d y}{d x} = \frac{- x e ^{- x^{2}}}{y}

What is the derivative of (x^2-4/(x+0.5)) ?
The derivative of (x^2-4/(x+0.5)) is (x^2+x+4)/((x+0.5)^2)
What is the first derivative of (x^2-4/(x+0.5)) ?
The first derivative of (x^2-4/(x+0.5)) is (x^2+x+4)/((x+0.5)^2)