Find the derivative of f(x) = x³sin(x²) + ln(x)cos(eˣ). Show all steps with detailed chain rule applications.

Let's find the derivative of the given function $f(x) = x^3\sin(x^2) + \ln(x)\cos(e^x)$.

We will use the sum rule, product rule, and chain rule for differentiation. The sum rule states that $\frac{d}{dx}[g(x) + h(x)] = \frac{d}{dx}[g(x)] + \frac{d}{dx}[h(x)]$. The product rule states that $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$. The chain rule states that $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$.

Step 1: Differentiate the first term, $x^3\sin(x^2)$. Let $u(x) = x^3$ and $v(x) = \sin(x^2)$. Then $u'(x) = \frac{d}{dx}[x^3] = 3x^2$. To find $v'(x)$, we apply the chain rule. Let $w = x^2$, so $v(x) = \sin(w)$. Then $\frac{dv}{dw} = \cos(w)$ and $\frac{dw}{dx} = 2x$. So, $v'(x) = \frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot (2x) = 2x\cos(x^2)$. Using the product rule for the first term: $\frac{d}{dx}[x^3\sin(x^2)] = u'(x)v(x) + u(x)v'(x)$ $= (3x^2)\sin(x^2) + (x^3)(2x\cos(x^2))$ $= 3x^2\sin(x^2) + 2x^4\cos(x^2)$

Step 2: Differentiate the second term, $\ln(x)\cos(e^x)$. Let $p(x) = \ln(x)$ and $q(x) = \cos(e^x)$. Then $p'(x) = \frac{d}{dx}[\ln(x)] = \frac{1}{x}$. To find $q'(x)$, we apply the chain rule. Let $z = e^x$, so $q(x) = \cos(z)$. Then $\frac{dq}{dz} = -\sin(z)$ and $\frac{dz}{dx} = e^x$. So, $q'(x) = \frac{d}{dx}[\cos(e^x)] = -\sin(e^x) \cdot (e^x) = -e^x\sin(e^x)$. Using the product rule for the second term: $\frac{d}{dx}[\ln(x)\cos(e^x)] = p'(x)q(x) + p(x)q'(x)$ $= \left(\frac{1}{x}\right)\cos(e^x) + (\ln(x))(-e^x\sin(e^x))$ $= \frac{\cos(e^x)}{x} - e^x\ln(x)\sin(e^x)$

Step 3: Combine the derivatives of both terms. $f'(x) = \frac{d}{dx}[x^3\sin(x^2)] + \frac{d}{dx}[\ln(x)\cos(e^x)]$ $f'(x) = (3x^2\sin(x^2) + 2x^4\cos(x^2)) + \left(\frac{\cos(e^x)}{x} - e^x\ln(x)\sin(e^x)\right)$ $f'(x) = 3x^2\sin(x^2) + 2x^4\cos(x^2) + \frac{\cos(e^x)}{x} - e^x\ln(x)\sin(e^x)$

The final answer is $\boxed{f'(x) = 3x^2\sin(x^2) + 2x^4\cos(x^2) + \frac{\cos(e^x)}{x} - e^x\ln(x)\sin(e^x)}$.