Power of Derivatives

$f(x) = e^x \cdot \sin x$

We can use the product rule of differentiation with $g(x) = e^x$ and $h(x) = \sin x$ respectively.

Thus, $f'(x) = g'(x) \cdot h(x) + h'(x) \cdot g(x) = e^x \cdot \sin x + \cos x \cdot e^x = e^x \cdot (\sin x + \cos x)$

Using the same method, we can proceed with the second degree differentiation:

$f''(x) = e^x \cdot (\sin x + \cos x) + (\cos x - \sin x) \cdot e^x = e^x \cdot (2\cos x) = 2e^x \cdot \cos x$

With the factor $2$ coefficient, the differentiation can continue with the factor remained:

$f'''(x) = 2e^x \cdot \cos x + 2e^x \cdot (-\sin x) = 2e^x \cdot (\cos x - \sin x)$

Finally, we can evaluate the fourth degree of differentiation:

$f^{(4)}(x) = 2e^x \cdot (\cos x - \sin x) + 2e^x \cdot (-\sin x - \cos x) = 2e^x \cdot (-2 \sin x) = -4e^x \cdot (\sin x)$

As we can see, the function $f^{(4)}(x) = -4 \cdot f(x)$ . In other words, by differentiation for $4$ times, such derivative can be evaluated by multiplying the original function $f$ by a factor of $-4$ .

Now if we differentiating for $2016$ times as stated in the question, the desired derivative can be calculated as shown:

$f^{(2016)} = (-4)^{\frac{2016}{4}} \cdot f(x) = 4^{504} \cdot f(x) = 2^{1008} \cdot f(x)$ .

Therefore, $\log_{2} \dfrac{f^{(2016)} (x)}{f(x)} = \log_{2} 2^{1008} = \boxed{1008}$

$\begin{aligned} f(x) & = e^x\sin x \\ & = \frac i2 \left(e^{(1-i)x} - e^{(1+i)x}\right) \\ f^{(1)} (x) & = \frac i2 \left((1-i)e^{(1-i)x} - (1+i)e^{(1+i)x}\right) \\ f^{(2)} (x) & = \frac i2 \left((1-i)^2e^{(1-i)x} - (1+i)^2e^{(1+i)x}\right) \\ f^{(3)} (x) & = \frac i2 \left((1-i)^3e^{(1-i)x} - (1+i)^3 e^{(1+i)x}\right) \\ ... & = \ ... \\ \implies f^{(2016)} (x) & = \frac i2 \left(\color{#3D99F6}{(1-i)^{2016}} e^{(1-i)x} - \color{#3D99F6}{(1+i)^{2016}} e^{(1+i)x}\right) & \small \color{#3D99F6}{\text{As }(1-i)^2 = -2i, \ (1+i)^2 = 2i } \\ & = \frac i2 \left(\color{#3D99F6}{(-2i)^{1008}} e^{(1-i)x} - \color{#3D99F6}{(2i)^{1008}} e^{(1+i)x}\right) & \small \color{#3D99F6}{\text{And } i^4 = 1} \\ & = \frac i2 \left(\color{#3D99F6}{(-2)^{1008}} e^{(1-i)x} - \color{#3D99F6}{(2)^{1008}} e^{(1+i)x}\right) \\ & = 2^{1008} \cdot \frac i2 \left(e^{(1-i)x} - e^{(1+i)x}\right) \\ & = 2^{1008} e^x \sin x \\ \implies \log_2 \frac {f^{(2016)}(x)}{f(x)} & = \boxed{1008} \end{aligned}$

By taking the first few derivatives of $f(x) = e^x\sin x$ , it can be observed that $f^{(4n)}(x) = \begin{cases} -2^{2n}f(x) \hspace{.2cm} \mbox{if} \hspace{.2cm} 4n \equiv 4 \mbox{(mod 8)}\\ \hspace{.3cm} 2^{2n}f(x) \hspace{.2cm} \mbox{if} \hspace{.2cm} 4n \equiv 8 \mbox{(mod 8)}\\ \end{cases}$

Since $8$ divides $2016$ , $f^{(2016)}(x) = f^{(4 \times 504)}(x) = 2^{1008}f(x)$

$\log_2 \left( \dfrac{f^{(2016)}(x)}{f(x)} \right) = \log_2 \left( \dfrac{2^{1008}f(x)}{f(x)} \right) = \log_2 \left(2^{1008}\right) = 1008$