What is the derivative of ((qx+r))/(sx+t) ?

The derivative of ((qx+r))/(sx+t) is x/(xs+t)

What is the first derivative of ((qx+r))/(sx+t) ?

The first derivative of ((qx+r))/(sx+t) is x/(xs+t)

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derivative of

\frac{( q x + r )}{s x + t}

\frac{x}{x s + t}

Solution steps

\frac{d}{d q} (\frac{q x + r}{s x + t})

Treat

x, r, s, t

as constants

Take the constant out:

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

= \frac{1}{s x + t} \frac{d}{d q} (q x + r)

Apply the Sum/Difference Rule:

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

= \frac{1}{s x + t} (\frac{d}{d q} (q x) + \frac{d r}{d q})

\frac{d}{d q} (q x) = x

\frac{d r}{d q} = 0

= \frac{1}{s x + t} (x + 0)

Simplify

\frac{1}{s x + t} (x + 0) : \frac{x}{x s + t}

= \frac{x}{x s + t}

\int_{0}^{1} 1 + (e^{x})^{2} d x

\int t^{n} d t

\int \frac{1}{( 3 - 5 x ) ^{4}} d x

\int 12 - 4 x d x

6 x - 4

What is the derivative of ((qx+r))/(sx+t) ?
The derivative of ((qx+r))/(sx+t) is x/(xs+t)
What is the first derivative of ((qx+r))/(sx+t) ?
The first derivative of ((qx+r))/(sx+t) is x/(xs+t)