What is the derivative of (-1/(2i+(y-x)j)) ?

The derivative of (-1/(2i+(y-x)j)) is (j\frac{dy)/(dx)-j}{(2i+j(y-x))^2}

What is the first derivative of (-1/(2i+(y-x)j)) ?

The first derivative of (-1/(2i+(y-x)j)) is (j\frac{dy)/(dx)-j}{(2i+j(y-x))^2}

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

cot (x^{- 2})

3 x^{2} y^{2}

\frac{e ^{2} x}{2}

lo g_{10} ((3 x + 2)^{3})

cos^{2} (\frac{x}{2})

What is the derivative of (-1/(2i+(y-x)j)) ?
The derivative of (-1/(2i+(y-x)j)) is (j\frac{dy)/(dx)-j}{(2i+j(y-x))^2}
What is the first derivative of (-1/(2i+(y-x)j)) ?
The first derivative of (-1/(2i+(y-x)j)) is (j\frac{dy)/(dx)-j}{(2i+j(y-x))^2}