Differentiate the following w.r.t. x:
`sin^(3)x+cos^(6)x`

Answer

To differentiate the function \( y = \sin^3 x + \cos^6 x \) with respect to \( x \), we will apply the chain rule and the power rule of differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate \( \sin^3 x \)
Using the chain rule, we differentiate \( \sin^3 x \):
\[ \frac{d}{dx}(\sin^3 x) = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) \]
Since \( \frac{d}{dx}(\sin x) = \cos x \), we have:
\[ \frac{d}{dx}(\sin^3 x) = 3 \sin^2 x \cdot \cos x \]

### Step 2: Differentiate \( \cos^6 x \)
Now, we differentiate \( \cos^6 x \) using the chain rule:
\[ \frac{d}{dx}(\cos^6 x) = 6 \cos^5 x \cdot \frac{d}{dx}(\cos x) \]
Since \( \frac{d}{dx}(\cos x) = -\sin x \), we get:
\[ \frac{d}{dx}(\cos^6 x) = 6 \cos^5 x \cdot (-\sin x) = -6 \cos^5 x \sin x \]

### Step 3: Combine the results
Now, we combine the results from Step 1 and Step 2 to find the derivative of \( y \):
\[ \frac{dy}{dx} = 3 \sin^2 x \cos x - 6 \cos^5 x \sin x \]

### Step 4: Factor the expression
We can factor out \( \sin x \) from the expression:
\[ \frac{dy}{dx} = \sin x (3 \sin x \cos x - 6 \cos^5 x) \]

### Final Result
Thus, the derivative of \( y = \sin^3 x + \cos^6 x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \sin x (3 \sin x \cos x - 6 \cos^5 x) \]

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