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

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Answer 1 · 2016-12-16T15:20:50

#(du)/(dt)=2^(t(t+1))ln2#

With functions like this use logarithmic differentiation.

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

#u=2^(t^2)#

#lnu=ln2^(t^2)#

using the laws of logs to simplify.

#lnu=t^2ln2#

differentiate with respect to #""t""#

#1/u(du)/(dt)=2tln2#

rearrange

#(du)/(dt)=u2tln2#

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

#(du)/(dt)=2^t2^(t^2)ln2#

#(du)/(dt)=2^(t^2+t)ln2#

#(du)/(dt)=2^(t(t+1))ln2#