\frac{d}{dt}\left(\sqrt{3t+\sqrt{2+\sqrt{1-t}}}\right)

asked by guest
on Nov 15, 2024 at 4:02 pm



You asked:

Evaluate the expression: ddt[3t+2+1−t]\frac{d}{dt}\left[\sqrt{3 t + \sqrt{2 + \sqrt{1 - t}}}\right]

MathBot Answer:

ddt[3t+2+1−t]=32−181−t1−t+23t+1−t+2\frac{d}{dt}\left[\sqrt{3 t + \sqrt{2 + \sqrt{1 - t}}}\right] = \frac{\frac{3}{2} - \frac{1}{8 \sqrt{1 - t} \sqrt{\sqrt{1 - t} + 2}}}{\sqrt{3 t + \sqrt{\sqrt{1 - t} + 2}}}

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