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

f (x) = - \frac{18}{x ^{3}}

\frac{d}{d x} (2 \cdot tan (x))

ln (6 x)

\frac{\partial}{\partial c} (16 d c^{2})

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