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Each of the following ``derivatives" has been done \textit{incorrectly}.  In each case below, find a counterexample that shows why the reasoning is incorrect.  Then correct the mistakes by finding the actual derivative of the left hand side.
\paragraph*{For example:}The statement ``$\frac{d}{dx} f(2x) =^{?} f'(2x)$" is false because if $f(x) = x^2$, then we have $\frac{d}{dx} f(2x) = \frac{d}{dx} [4x^2] = 8x$, but on the other hand, $f'(x) = 2x$, so $f'(2x) = 4x$.  In general, the correct derivative would be ``$\frac{d}{dx} f(2x) = f'(2x) \cdot 2$" by the chain rule.
\begin{enumerate}
\item[(a)]  $\dfrac{d}{dx} f(x+5) =^{?} f'(x)$
\item[(b)]  $\dfrac{d}{dt} g(t/3) =^{?} g'(t)/3$
\item[(c)]  $\dfrac{d}{ds} h(s^2) =^{?} h'(2s)$
\item[(d)]  $\dfrac{d}{dt} k(x) =^{?} k'(x)$
\item[(f)]  $\dfrac{d}{dx} l(x\cos(x)) =^{?} l(\cos(x) - x\sin(x)) \cdot x \cos(x)$
\item[(g)]  $\dfrac{d}{du} \left[ F(u)\sin(2u) \right] =^{?} 2F'(u)\cos(2u)$
\item[(h)]  $\dfrac{d}{dt} G(\sin(t)) =^{?} G'(\sin(t)) + \cos(t)$
\end{enumerate}

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