Answer is done, 

First note that $f$ has to be continuous. Hence we should have 
\[\lim_{x\rightarrow 4^-}f(x)=\lim_{x\rightarrow 4^+}f(x)   \Rightarrow  c(4)^2=10-4d.\]
Thus
\[16c=10-4d.       (1)\]


For $x< 4$, we have
\[f'(x)=2cx    \Rightarrow  \lim_{x\rightarrow 4^-}f'(x)=2c(4)=8c.\]

For $x>4$ we have

\[f'(x)=-d    \Rightarrow  \lim_{x\rightarrow 4^+}f'(x)=-d.\]

For $f'$ to be continuous, we should have 

\[\lim_{x\rightarrow 4^-}f'(x)=\lim_{x\rightarrow 4^+}f'(x)   \Rightarrow  8c=-d.\]
Thus
\[8c=-d.     (2)\]
Solving the symstem of equations (1) and (2), we have
\[16c=10-4d,     16c=-2d \Rightarrow  10-4d=-2d\]
\[10=2d   \Rightarrow   d=5. \]
Also 
\[16c=-2(5)=-10. \Rightarrow c=-\frac{-10}{16}=-\frac{5}{8}.\]