What is the derivative of 2/(xy) ?

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

What is the first derivative of 2/(xy) ?

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

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\frac{d}{d x} (\frac{2}{x y})

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

Solution steps

\frac{d}{d x} (\frac{2}{x y})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{\frac{π}{4}} 2 sec (x) d x

f^{^{'}} (x) = x^{2} e^{x^{2}}

\frac{3 x}{x + 1}

2 x^{2} - \frac{1}{16} ln (x)

\int cos^{3} (x) sin^{6} (x) d x

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