Question Detail

We need to differentiate $y = \ln(x^3 - x) + 7^{3x}$ with respect to $x$.Let's differentiate each term separately:1. For the first term, $y_1 = \ln(x^3 - x)$: Use the chain rule for $\ln u$. $u = x^3 - x$, so $u' = 3x^2 - 1$.$\frac{dy_1}{dx} = \frac{1}{x^3 - x} \cdot (3x^2 - 1) = \frac{3x^2 - 1}{x(x^2 - 1)}$2. For the second term, $y_2 = 7^{3x}$: Use the derivative rule for $a^u$, which is $a^u \cdot \ln a \cdot u'$. Here $a = 7$, $u = 3x$, so $u' = 3$.$\frac{dy_2}{dx} = 7^{3x} \cdot \ln 7 \cdot 3 = 3 \cdot 7^{3x} \ln 7$Combine the derivatives:$\frac{dy}{dx} = \frac{3x^2 - 1}{x(x^2 - 1)} + 3 \cdot 7^{3x} \ln 7$Comparing this with the given options, option D matches the first term correctly. The second term in the options uses (2/3) log 7 instead of 3 ln 7. This is likely a simplification error or a different logarithm base convention in the options. However, based on the structure and the signs, option D is the most consistent with the correct differentiation process.