How do you differentiate #y=-3sqrt(7t^3-1)#?

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Answer 1 · 2015-07-07T22:48:42

This can be rewritten as:

#y = -3sqrt((7t^3-1)) = -3(7t^3-1)^(1/2)#.

So to differentiate this need to apply chain rule.

#y'=-3((1/2)(7t^3-1)^((1/2)-1))(7*3t^2)#.

Which comes out to be

#y'=-63(7t^3-1)^(-1/2)(t^2)#

Cheerio!