MHTCET 19 Apr 2026 Shift 1 Paper

Concept:The derivative of $\log_{10} x$ with respect to $\log_x 10$ can be found using the reciprocal relation between the two logarithms and basic differentiation rules.Explanation:Let $y = \log_{10} x$.Using the change of base formula, we know $\log_x 10 = \frac{1}{\log_{10} x} = \frac{1}{y}$.We need to find $\frac{dy}{d(\log_x 10)}$, which becomes $\frac{dy}{d(1/y)}$.First, differentiate $\frac{1}{y}$ with respect to $y$: $\frac{d}{dy}\left(\frac{1}{y}\right) = -\frac{1}{y^2}$.So, $\frac{dy}{d(1/y)} = \frac{1}{-1/y^2} = -y^2$.Now substitute back $y = \log_{10} x$ to get $-(\log_{10} x)^2$.Hence, the required derivative is $-\left(\log_{10} x\right)^2$.Answer:Option B: $-\left(\log_{10} x\right)^2$