What is the derivative of (x+yi-(x^2+y^2)j) ?

The derivative of (x+yi-(x^2+y^2)j) is (-2jx-2y(dy)/(dx))+i(1+(dy)/(dx))

What is the first derivative of (x+yi-(x^2+y^2)j) ?

The first derivative of (x+yi-(x^2+y^2)j) is (-2jx-2y(dy)/(dx))+i(1+(dy)/(dx))

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

(- 2 j x - 2 y \frac{d y}{d x}) + i (1 + \frac{d y}{d x})

Solution steps

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

Simplify

(x + y) i - (x^{2} + y^{2}) j : (- j x^{2} - j y (x)^{2}) + i (y + x)

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Rewrite

- 2 j x - 2 y \frac{d y}{d x} + i (\frac{d y}{d x} + 1)

in standard complex form:

(- 2 j x - 2 y \frac{d y}{d x}) + (\frac{d y}{d x} + 1) i

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

lo g_{10} (sin (x - x^{0}))

x^{2} + y^{2} - 6 x + 8 y

\frac{a ^{2}}{x ^{2}}

- 4 e^{- 2} x

v (t) = (1 - t) (1 + t^{2})^{- 1}

What is the derivative of (x+yi-(x^2+y^2)j) ?
The derivative of (x+yi-(x^2+y^2)j) is (-2jx-2y(dy)/(dx))+i(1+(dy)/(dx))
What is the first derivative of (x+yi-(x^2+y^2)j) ?
The first derivative of (x+yi-(x^2+y^2)j) is (-2jx-2y(dy)/(dx))+i(1+(dy)/(dx))