What is the d/(dq)(({r}(q))(60-0.12q)) ?

The d/(dq)(({r}(q))(60-0.12q)) is d/(dq)({r}(q))(60-0.12q)-0.12{r}(q)

What is the first d/(dq)(({r}(q))(60-0.12q)) ?

The first d/(dq)(({r}(q))(60-0.12q)) is d/(dq)({r}(q))(60-0.12q)-0.12{r}(q)

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\frac{d}{d q} ((r (q)) (60 - 0.12 q))

\frac{d}{d q} (r (q)) (60 - 0.12 q) - 0.12 r (q)

Solution steps

\frac{d}{d q} ((r (q)) (60 - 0.12 q))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = r (q), g = 60 - 0.12 q

= \frac{d}{d q} (r (q)) (60 - 0.12 q) + \frac{d}{d q} (60 - 0.12 q) r (q)

\frac{d}{d q} (60 - 0.12 q) = - 0.12

= \frac{d}{d q} (r (q)) (60 - 0.12 q) + (- 0.12) r (q)

Simplify

= \frac{d}{d q} (r (q)) (60 - 0.12 q) - 0.12 r (q)

\frac{( y - x ) d x + y}{d x}

100 (1 - \frac{1}{e ^{2}})

\frac{8 x ( x - 5 )}{4 x ^{2} - 27 x - 7}

8 (2 x + 3 y = 15)

3 - 3 \frac{3}{4}

What is the d/(dq)(({r}(q))(60-0.12q)) ?
The d/(dq)(({r}(q))(60-0.12q)) is d/(dq)({r}(q))(60-0.12q)-0.12{r}(q)
What is the first d/(dq)(({r}(q))(60-0.12q)) ?
The first d/(dq)(({r}(q))(60-0.12q)) is d/(dq)({r}(q))(60-0.12q)-0.12{r}(q)