Question

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

Answer 1

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:

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