` If y= cos (x^(2) e^(x) ),then (dy)/(dx) =`

Options

A. ` e^(x) (x^(2)+2)sin (x^(2) e^(x)) `
B. `- e^(x) (x^(2)+2)sin (x^(2) e^(x)) `
C. `x e^(x) (x+2)sin (x^(2) e^(x)) `
D. `-x e^(x) (x+2)sin (x^(2) e^(x)) `

Answer

To find the derivative of the function \( y = \cos(x^2 e^x) \), we will use the chain rule and the product rule. Here’s the step-by-step solution:

### Step 1: Identify the outer function and the inner function
Let \( u = x^2 e^x \). Then, we can rewrite \( y \) as:
\[ y = \cos(u) \]

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( y \) with respect to \( u \) is:
\[ \frac{dy}{du} = -\sin(u) \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = x^2 e^x \). We will apply the product rule here. The product rule states that if you have two functions \( f(x) \) and \( g(x) \), then:
\[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]
For our case, let \( f(x) = x^2 \) and \( g(x) = e^x \).

First, we find the derivatives:
- \( f'(x) = 2x \)
- \( g'(x) = e^x \)

Now applying the product rule:
\[ \frac{du}{dx} = f'(x)g(x) + f(x)g'(x) = (2x)(e^x) + (x^2)(e^x) \]
This simplifies to:
\[ \frac{du}{dx} = e^x(2x + x^2) \]

### Step 4: Combine the derivatives using the chain rule
Now we can use the chain rule to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\sin(u) \cdot e^x(2x + x^2) \]
Substituting back \( u = x^2 e^x \):
\[ \frac{dy}{dx} = -\sin(x^2 e^x) \cdot e^x(2x + x^2) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -e^x(2x + x^2) \sin(x^2 e^x) \]