1. Stated problem: Differentiate $T(z) = 4^z \log_4(z)$. 2. Recall the product rule for derivatives: if $T(z) = u(z)v(z)$, then $$T'(z) = u'(z)v(z) + u(z)v'(z)$$ 3. Define the functions: $u(z) = 4^z$ $v(z) = \log_4(z)$ 4. Differentiate $u(z)$ using exponentials and natural logs: $$u(z) = 4^z = e^{z \ln(4)}$$ $$u'(z) = \frac{d}{dz} e^{z \ln(4)} = e^{z \ln(4)} \ln(4) = 4^z \ln(4)$$ 5. Differentiate $v(z)$ by converting base 4 logarithm to natural log: $$v(z) = \log_4(z) = \frac{\ln(z)}{\ln(4)}$$ $$v'(z) = \frac{1}{z \ln(4)}$$ 6. Substitute derivatives into product rule: $$T'(z) = u'(z)v(z) + u(z)v'(z) = 4^z \ln(4) \cdot \frac{\ln(z)}{\ln(4)} + 4^z \cdot \frac{1}{z \ln(4)}$$ 7. Simplify terms: $$T'(z) = 4^z \ln(z) + \frac{4^z}{z \ln(4)}$$ Final answer: $$\boxed{T'(z) = 4^z \ln(z) + \frac{4^z}{z \ln(4)}}$$