MHT CET · Maths · Differentiation
If \(\mathrm{y}=a^x \cdot \mathrm{~b}^{2 x-1}\), then \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}\) is equal to
- A \(\mathrm{y}\left(\log \left(a \mathrm{~b}^2\right)\right)\)
- B \(\mathrm{y}^2\left(\log \left(a \mathrm{~b}^2\right)\right)\)
- C \(\mathrm{y}\left(\log \left(a \mathrm{~b}^2\right)\right)^2\)
- D \(y^2(\log (a b))^2\)
Answer & Solution
Correct Answer
(C) \(\mathrm{y}\left(\log \left(a \mathrm{~b}^2\right)\right)^2\)
Step-by-step Solution
Detailed explanation
\(y = a^x \cdot b^{2x-1} = a^x \cdot b^{2x} \cdot b^{-1} = (a b^2)^x \cdot b^{-1}\) \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} \left( (a b^2)^x \cdot b^{-1} \right) = (a b^2)^x \cdot \ln(a b^2) \cdot b^{-1} = y \ln(a b^2)\)
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