Derivative of `cos x^3 sin^2(x^5)` with respect to x is:

Options

A. `10x^4(sin x^5)(cosx^5)(cosx^3)-3x^2sinx^3sin^2x^5`
B. `5x^3(sin x^5)(cosx^5)(cosx^3)-3x^2sinx^3sin^2x^5`
C. `5x^3(sin x^5)(cosx^5)(cosx^3)-6x^2sinx^3sin^2x^5`
D. None of the above

Answer

To find the derivative of \( y = \cos(x^3) \sin^2(x^5) \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if \( y = u \cdot v \), then the derivative \( y' \) is given by:

\[ y' = u'v + uv' \]

where \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \cos(x^3) \)
- \( v = \sin^2(x^5) \)

### Step 2: Differentiate \( u \)
To find \( u' \):
\[ u' = \frac{d}{dx}[\cos(x^3)] \]
Using the chain rule:
\[ u' = -\sin(x^3) \cdot \frac{d}{dx}(x^3) = -\sin(x^3) \cdot 3x^2 \]
Thus,
\[ u' = -3x^2 \sin(x^3) \]

### Step 3: Differentiate \( v \)
To find \( v' \):
\[ v = \sin^2(x^5) = (\sin(x^5))^2 \]
Using the chain rule:
\[ v' = 2\sin(x^5) \cdot \frac{d}{dx}[\sin(x^5)] \]
Now, differentiate \( \sin(x^5) \):
\[ \frac{d}{dx}[\sin(x^5)] = \cos(x^5) \cdot \frac{d}{dx}(x^5) = \cos(x^5) \cdot 5x^4 \]
So,
\[ v' = 2\sin(x^5) \cdot (5x^4 \cos(x^5)) = 10x^4 \sin(x^5) \cos(x^5) \]

### Step 4: Apply the Product Rule
Now, we can substitute \( u, u', v, \) and \( v' \) into the product rule formula:
\[ y' = u'v + uv' \]
Substituting the values:
\[ y' = (-3x^2 \sin(x^3)) \cdot (\sin^2(x^5)) + (\cos(x^3)) \cdot (10x^4 \sin(x^5) \cos(x^5)) \]

### Step 5: Simplify the Expression
Now we simplify:
\[ y' = -3x^2 \sin(x^3) \sin^2(x^5) + 10x^4 \sin(x^5) \cos(x^5) \cos(x^3) \]

### Final Answer
Thus, the derivative of \( \cos(x^3) \sin^2(x^5) \) with respect to \( x \) is:
\[ y' = 10x^4 \sin(x^5) \cos(x^5) \cos(x^3) - 3x^2 \sin(x^3) \sin^2(x^5) \]

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