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MHT CET 2025 21st April Morning Shift
MCQ (More than One Correct Answer)
+2
-0

If $x=\log \mathrm{t}, \mathrm{t}>0$ and $y=\frac{1}{\mathrm{t}}$ then $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=$

A
$\frac{\mathrm{d} y}{\mathrm{~d} x}$
B
$-\frac{\mathrm{d} y}{\mathrm{~d} x}$
C
$y$
D
$\frac{y}{x}$
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