Exercise

Correct: Write $c(q) = u(q) + v(q)$ with $u(q) = 20\sqrt{q} = 20q^{\tfrac{1}{2}}$ and $v(q) = 5q + 100$. Then $v(q)$ is again the sum of two functions: $v(q) = w(q) + y(q)$, with $w(q) = 5q$ and $y(q) = 100$. By the use of the Derivatives of elementary functions and the Scalar product rule we then find $c'(q)$:
$$\begin{align*} u'(q) &= 20 \cdot \tfrac{1}{2} q^{\tfrac{1}{2}-1} = 10q^{-\tfrac{1}{2}} = \dfrac{10}{\sqrt{q}}\\ w'(q) &= 5\\ y'(q) &= 0\\ v'(q) &= w'(q) + y'(q) = 5\\ c'(q) &= u'(q) + v'(q) = \dfrac{10}{\sqrt{q}} + 5. \end{align*}$$
Finally, we plug in $q=4$:
$$c'(4) = \dfrac{10}{\sqrt{4}} + 5= \dfrac{10}{2} + 5 = 5+ 5 = 10.$$

Go on.

Wrong: You have to calculate the derivative in the point $q=4$, not the derivative function $c'(q)$.

Try again.

Wrong: The derivative of $\sqrt{q}$ can be determined.

See Properties power functions or Derivatives of elementary functions: Example 2.

Wrong: Write $c(q) = u(q) + v(q)$, with $u(q)=20\sqrt{q}$. Then write $v(q) = w(q) + y(q)$ and apply the sum rule twice (once to determine $v'(q)$ and once to determine $c'(q)$).

Try again.