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If $t=e^{2 x}$ and $y=\log _e t^2$, then $\frac{d^2 y}{d x^2}$ is: |
0 4t $\frac{4e^{2t}}{t}$ $\frac{e^{2t}(4t-1)}{t^2}$ |
0 |
The correct answer is Option (1) → 0 $t=e^{2 x}$, $y=\log _e t^2$ so $\log t=2x$ so $\frac{\log t}{2}=x,y=2\log t$ so $\frac{dx}{dt}=\frac{1}{2t}$, $\frac{dy}{dt}=\frac{2}{t}$ $\frac{dy}{dx}=4⇒\frac{d^2y}{dx^2}=0$ |
