How do you find the derivative of $u = 2^{t^{2}}$ $u=2^(t^2)$ ?

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Answer 1 · 2016-12-16T15:20:50

$\frac{d u}{d t} = 2^{t (t + 1)} ln 2$ $(du)/(dt)=2^(t(t+1))ln2$

With functions like this use logarithmic differentiation.

Take logs to base $e$ $""e""$ first, then differentiate.

$u = 2^{t^{2}}$ $u=2^(t^2)$

$ln u = {ln 2}^{t^{2}}$ $lnu=ln2^(t^2)$

using the laws of logs to simplify.

$ln u = t^{2} ln 2$ $lnu=t^2ln2$

differentiate with respect to $t$ $""t""$

$\frac{1}{u} \frac{d u}{d t} = 2 t ln 2$ $1/u(du)/(dt)=2tln2$

rearrange

$\frac{d u}{d t} = u 2 t ln 2$ $(du)/(dt)=u2tln2$

substitute back for $u$ $""u""$ and tidy up.

$\frac{d u}{d t} = 2^{t} 2^{t^{2}} ln 2$ $(du)/(dt)=2^t2^(t^2)ln2$

$\frac{d u}{d t} = 2^{t^{2} + t} ln 2$ $(du)/(dt)=2^(t^2+t)ln2$

$\frac{d u}{d t} = 2^{t (t + 1)} ln 2$ $(du)/(dt)=2^(t(t+1))ln2$

How do you find the derivative of u=2^(t^2)u=2t2u=2^(t^2)?