If `y=x^(3)cosx" then "(dy)/(dx)`=……………………

Options

A. `x^(2)(3cosx-xsinx)`
B. `x^(2)(3cosx+xsinx)`
C. `3x^(2).cosx+x^(3)sinx`
D. None of these

Answer

To find the derivative of the function \( y = x^3 \cos x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u'v + uv' \]

### Step 1: Identify the functions
Let:
- \( u = x^3 \)
- \( v = \cos x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find the derivatives of \( u \) and \( v \):
- The derivative of \( u \) is:
\[ u' = \frac{d}{dx}(x^3) = 3x^2 \]
- The derivative of \( v \) is:
\[ v' = \frac{d}{dx}(\cos x) = -\sin x \]

### Step 3: Apply the product rule
Now we apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (3x^2)(\cos x) + (x^3)(-\sin x) \]

### Step 4: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = 3x^2 \cos x - x^3 \sin x \]

### Step 5: Factor the expression (if necessary)
We can factor out \( x^2 \):
\[ \frac{dy}{dx} = x^2(3 \cos x - x \sin x) \]

Thus, the final answer is:
\[ \frac{dy}{dx} = x^2(3 \cos x - x \sin x) \]

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