What is the first derivative of qx+r ?

The first derivative of qx+r is x

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derivative of qx+r

derivative of

q x + r

x

Solution steps

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

Treat

r, x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= x + 0

Simplify

= x

d er i v a t i v e f (x) = - \frac{18}{x ^{3}}

d er i v a t i v e ln (6 x)

\frac{d y}{d x} + \frac{y}{x} = 3 x^{3} y^{2}

\int 3 sec^{2} (x) + 2 d x

3 x^{2} y^{2} d x + (2 x^{3} y + 4 y^{3}) d y = 0