Question

Find the Derivative \( \frac{dy}{dx} \)
Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]

Answer 1

Step 1: Differentiate term-by-term

Given: \[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \] Using the sum rule: \[ \frac{dy}{dx} = \frac{d}{dx}[\cos(x^2)] + \frac{d}{dx}[\cos(2x)] + \frac{d}{dx}[\cos^2(x^2)] + \frac{d}{dx}[\cos(x^x)] \]

Step 2: First term \(\cos(x^2)\)

Using chain rule: \[ \frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -\sin(x^2) \cdot 2x \]

Step 3: Second term \(\cos(2x)\)

Again chain rule: \[ \frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot 2 \]

Step 4: Third term \(\cos^2(x^2)\)

Let \(u = \cos(x^2)\), then the term is \(u^2\): \[ \frac{d}{dx}[\cos^2(x^2)] = 2\cos(x^2) \cdot \frac{d}{dx}[\cos(x^2)] \] We already know: \[ \frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot 2x \] So: \[ \frac{d}{dx}[\cos^2(x^2)] = 2\cos(x^2) \cdot [-\sin(x^2) \cdot 2x] = -4x\cos(x^2)\sin(x^2) \]

Step 5: Fourth term \(\cos(x^x)\)

Let \(v = x^x\). We know: \[ \frac{d}{dx}(x^x) = x^x \left( \ln x + 1 \right) \] Now apply chain rule: \[ \frac{d}{dx}[\cos(x^x)] = -\sin(x^x) \cdot \frac{d}{dx}(x^x) = -\sin(x^x) \cdot x^x(\ln x + 1) \]

Step 6: Combine all results

\[ \frac{dy}{dx} = -2x\sin(x^2) - 2\sin(2x) - 4x\cos(x^2)\sin(x^2) - x^x(\ln x + 1)\sin(x^x) \]

Final Answer:
\[ \boxed{\frac{dy}{dx} = -2x\sin(x^2) - 2\sin(2x) - 4x\cos(x^2)\sin(x^2) - x^x(\ln x + 1)\sin(x^x)} \]

Answer 2

Term 1: \( \frac{d}{dx}[\cos(x^2)] \)

Using chain rule:

\[ \frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -2x \sin(x^2) \]

Term 2: \( \frac{d}{dx}[\cos^2(x)] \)

\( \cos^2(x) = (\cos x)^2 \), use chain rule:

\[ \frac{d}{dx}[\cos^2(x)] = 2\cos(x) \cdot (-\sin(x)) = -2\cos(x)\sin(x) \]

Term 3: \( \frac{d}{dx}[\cos^2(x^2)] \)

Let \( u = x^2 \), then:

\[ \frac{d}{dx}[\cos^2(x^2)] = 2\cos(x^2)(-\sin(x^2)) \cdot \frac{d}{dx}(x^2) = -4x \cos(x^2)\sin(x^2) \]

Term 4: \( \frac{d}{dx}[\cos(x^x)] \)

Let \( u = x^x \). Then:

\[ \frac{d}{dx}[x^x] = \frac{d}{dx}[e^{x\ln x}] = e^{x\ln x} \cdot \frac{d}{dx}(x\ln x) = x^x(\ln x + 1) \] \[ \frac{d}{dx}[\cos(x^x)] = -\sin(x^x) \cdot \frac{d}{dx}(x^x) = -x^x(\ln x + 1)\sin(x^x) \]

✅ Final Answer:

\[ \frac{dy}{dx} = -2x \sin(x^2) - 2 \cos(x) \sin(x) - 4x \cos(x^2) \sin(x^2) - x^x (\ln x + 1) \sin(x^x) \]