`(x^(3) cosx - 2^(x) tanx)`

Answer

To differentiate the expression \( y = x^3 \cos x - 2^x \tan x \), we will apply the product rule and the chain rule where necessary. Here’s a step-by-step solution:

### Step 1: Identify the functions
We have two main parts to differentiate:
1. \( u = x^3 \) and \( v = \cos x \) for the first term \( x^3 \cos x \)
2. \( a = 2^x \) and \( b = \tan x \) for the second term \( 2^x \tan x \)

### Step 2: Differentiate the first term \( x^3 \cos x \)
Using the product rule:
\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]
where:
- \( u = x^3 \) and \( \frac{du}{dx} = 3x^2 \)
- \( v = \cos x \) and \( \frac{dv}{dx} = -\sin x \)

So,
\[ \frac{d}{dx}(x^3 \cos x) = x^3(-\sin x) + \cos x(3x^2) \]
This simplifies to:
\[ - x^3 \sin x + 3x^2 \cos x \]

### Step 3: Differentiate the second term \( 2^x \tan x \)
Again using the product rule:
\[ \frac{d}{dx}(ab) = a \frac{db}{dx} + b \frac{da}{dx} \]
where:
- \( a = 2^x \) and \( \frac{da}{dx} = 2^x \ln 2 \)
- \( b = \tan x \) and \( \frac{db}{dx} = \sec^2 x \)

So,
\[ \frac{d}{dx}(2^x \tan x) = 2^x \sec^2 x + \tan x(2^x \ln 2) \]
This simplifies to:
\[ 2^x \sec^2 x + 2^x \tan x \ln 2 \]

### Step 4: Combine the derivatives
Now we can combine the results from Steps 2 and 3:
\[ \frac{dy}{dx} = \left(- x^3 \sin x + 3x^2 \cos x\right) - \left(2^x \sec^2 x + 2^x \tan x \ln 2\right) \]
This gives us:
\[ \frac{dy}{dx} = - x^3 \sin x + 3x^2 \cos x - 2^x \sec^2 x - 2^x \tan x \ln 2 \]

### Final Answer
Thus, the derivative of the function \( y = x^3 \cos x - 2^x \tan x \) is:
\[ \frac{dy}{dx} = - x^3 \sin x + 3x^2 \cos x - 2^x \sec^2 x - 2^x \tan x \ln 2 \]