Question

Differentiate 4^xlnx

Answer 1

If you mean

$\dfrac{\mathrm d}{\mathrm dx}\left[4^x\ln(x)\right]$

then first write $4^x=e^{\ln(4^x)}=e^{\ln(4)x}$. Then by the product rule,

$\dfrac{\mathrm d}{\mathrm dx}\left[4^x\ln(x)\right] = \ln(4)e^{\ln(4)x}\ln(x) + \dfrac{e^{\ln(4)x}}x = \ln(4)4^x\ln(x)+\dfrac{4^x}x=\dfrac{4^x}x\left(\ln(4)x\ln(x)+1\right)$

If you instead mean

$\dfrac{\mathrm d}{\mathrm dx}\left[4^{x\ln(x)}\right]$

you can again rewrite $4^{x\ln(x)} = e^{\ln\left(4^{x\ln(x)}\right)}=e^{x\ln(x)\ln(4)}$. Then by the chain and product rules,

$\dfrac{\mathrm d}{\mathrm dx}\left[4^{x\ln(x)}\right] = e^{x\ln(x)\ln(4)}\ln(4)\left(\ln(x)+1\right) = 4^{x\ln(x)}\ln(4)(\ln(x)+1)$