What is the derivative of (kx+a/(bx+c)) ?

The derivative of (kx+a/(bx+c)) is (-ab+ck)/((bx+c)^2)

What is the first derivative of (kx+a/(bx+c)) ?

The first derivative of (kx+a/(bx+c)) is (-ab+ck)/((bx+c)^2)

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\frac{d}{d x} (\frac{k x + a}{b x + c})

\frac{- ab + c k}{( b x + c ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{k x + a}{b x + c})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( k x + a ) ( b x + c ) - \frac{d}{d x} ( b x + c ) ( k x + a )}{( b x + c ) ^{2}}

\frac{d}{d x} (k x + a) = k

\frac{d}{d x} (b x + c) = b

= \frac{k ( b x + c ) - b ( k x + a )}{( b x + c ) ^{2}}

Expand

k (b x + c) - b (k x + a) : - ab + c k

= \frac{- ab + c k}{( b x + c ) ^{2}}

What is the derivative of (kx+a/(bx+c)) ?
The derivative of (kx+a/(bx+c)) is (-ab+ck)/((bx+c)^2)
What is the first derivative of (kx+a/(bx+c)) ?
The first derivative of (kx+a/(bx+c)) is (-ab+ck)/((bx+c)^2)