Question

Differentiate: sqrt{2+t^2}
u = 2 + t^2
dt = du/2t
int_{}{}u ^{1/2} dt
(u ^{3/2})/(3/2) * du/2t

Answer: (2+t^2)^{3/2} / 3t
Is it correct?

Answer 1

\[2t*1/(\sqrt{2+t^2})\]

Answer 2

\[\Large \sqrt{2+t^2}=(2+t^2)^{\frac{1}{2}}\]


\[\Large \frac{d}{dx}\left(\sqrt{2+t^2}\right)=\frac{d}{dx}\left((2+t^2)^{\frac{1}{2}}\right)\]


\[\Large \frac{d}{dx}\left(\sqrt{2+t^2}\right)=\frac{1}{2}(2+t^2)^{-\frac{1}{2}}\times \frac{d}{dx}(2+t^2)\]


\[\Large \frac{d}{dx}\left(\sqrt{2+t^2}\right)=\frac{2t}{2\sqrt{2+t^2}}\]


\[\Large \frac{d}{dx}\left(\sqrt{2+t^2}\right)=\frac{t}{\sqrt{2+t^2}}\]

Answer 3

Corrected formatting:

\[\sqrt{2+t^2}\]
\[u = 2 + t^2\]
\[dt = du/2t\]
\[\int\limits\limits_{}{}u ^{1/2} dt\]
\[(u ^{3/2})/(3/2) * du/2t \]
\[Answer: (2+t^2)^{3/2} / 3t\]

Answer 4

oops