What is the derivative of (Ax+B/(Cx+d)) ?

The derivative of (Ax+B/(Cx+d)) is (Ad-BC)/((xC+d)^2)

What is the first derivative of (Ax+B/(Cx+d)) ?

The first derivative of (Ax+B/(Cx+d)) is (Ad-BC)/((xC+d)^2)

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\frac{d}{d x} (\frac{A x + B}{C x + d})

\frac{A d - BC}{( x C + d ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{A x + B}{C x + d})

Treat

A, B, C, d

as constants

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( A x + B ) ( C x + d ) - \frac{d}{d x} ( C x + d ) ( A x + B )}{( C x + d ) ^{2}}

\frac{d}{d x} (A x + B) = A

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

= \frac{A ( C x + d ) - C ( A x + B )}{( C x + d ) ^{2}}

Expand

A (C x + d) - C (A x + B) : A d - BC

= \frac{A d - BC}{( x C + d ) ^{2}}

What is the derivative of (Ax+B/(Cx+d)) ?
The derivative of (Ax+B/(Cx+d)) is (Ad-BC)/((xC+d)^2)
What is the first derivative of (Ax+B/(Cx+d)) ?
The first derivative of (Ax+B/(Cx+d)) is (Ad-BC)/((xC+d)^2)