To solve: $\frac{d}{dx}(-3 \cdot 2^x)$, do I need all the other rules? Or will the derivative of $a^{x}$(for positive base a) be sufficient?
Here's how I did it:
$\frac{d}{dx}(-3 \cdot 2^x) = F(x) = f(g(x))$
- Apply Chain Rule
$\begin{aligned} F(x)=f^{\prime}(x) &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \\ &=\frac{d}{d x}(-3 \cdot g(x)) \cdot \frac{d}{d x}\left(2^{x}\right) \end{aligned}$
- Apply Constant rule
- Apply Power rule
- Apply Derivative of $a^{x}$(for positive base a)
$\begin{aligned}=&-3 \frac{d}{d x}(1(g(x))^0) \cdot \ln (2) 2^{x} \\=&-3\left(1 \times\left(2^{x}\right)^0\right) \cdot \ln (2) 2^{x} \\ &=-3 \ln (2) 2^{x} \text { or } 2^{x} \cdot-3 \ln (2) \end{aligned}$
However, in Khan acad the solution is 3 steps.
I'm a self-taught student, so could use any guidance if there are short cuts and conventions, thank you!