What is the d/(dq)(q-r+s) ?

The d/(dq)(q-r+s) is -(dr)/(dq)+1

What is the first d/(dq)(q-r+s) ?

The first d/(dq)(q-r+s) is -(dr)/(dq)+1

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\frac{d}{d q} (q - r + s)

- \frac{d r}{d q} + 1

Solution steps

\frac{d}{d q} (q - r + s)

Apply the Sum/Difference Rule:

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

= \frac{d q}{d q} - \frac{d r}{d q} + \frac{d s}{d q}

\frac{d q}{d q} = 1

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

= 1 - \frac{d r}{d q} + 0

Simplify

= - \frac{d r}{d q} + 1

- 5 s = - 13

(\frac{1}{2})^{- 3}

4 x^{2} + 28 x = 0

\frac{42 y z ^{4}}{- 48 x z ^{3}}

- (\frac{5}{2})^{- 2}