What is the derivative of ((x^3+ix^2+x+1)/(x^2+i)) ?

The derivative of ((x^3+ix^2+x+1)/(x^2+i)) is (x^4+3ix^2-x^2-4x+i)/((x^2+i)^2)

What is the first derivative of ((x^3+ix^2+x+1)/(x^2+i)) ?

The first derivative of ((x^3+ix^2+x+1)/(x^2+i)) is (x^4+3ix^2-x^2-4x+i)/((x^2+i)^2)

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

\frac{x ^{4} + 3 i x ^{2} - x ^{2} - 4 x + i}{( x ^{2} + i ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{3} + i x^{2} + x + 1) = (3 x^{2} + 1) + 2 i x

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

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

Simplify

\frac{( 3 x ^{2} + 1 + 2 i x ) ( x ^{2} + i ) - 2 x ( x ^{3} + i x ^{2} + x + 1 )}{( x ^{2} + i ) ^{2}} : \frac{x ^{4} + 3 i x ^{2} - x ^{2} - 4 x + i}{( x ^{2} + i ) ^{2}}

= \frac{x ^{4} + 3 i x ^{2} - x ^{2} - 4 x + i}{( x ^{2} + i ) ^{2}}

\int (1 - y) d y

\frac{d y}{d t} + 0.3 t y = 5 t, y (0) = 6

x \to 3 + lim (\frac{8}{x - 3})

\frac{d}{d x} (2 sin (x) - cos (x))

(e^{- 3 y}) y^{^{'}} = e^{8 x}, y (0) = 12

What is the derivative of ((x^3+ix^2+x+1)/(x^2+i)) ?
The derivative of ((x^3+ix^2+x+1)/(x^2+i)) is (x^4+3ix^2-x^2-4x+i)/((x^2+i)^2)
What is the first derivative of ((x^3+ix^2+x+1)/(x^2+i)) ?
The first derivative of ((x^3+ix^2+x+1)/(x^2+i)) is (x^4+3ix^2-x^2-4x+i)/((x^2+i)^2)