How do you use the definition of a derivative to find the derivative of #G(t) = (1-6t)/(5+t)#?

1 Answer
Steve M
Dec 13, 2016

# G'(t) = (-31 ) / (5+t)^2#

Explanation:

The definition of the derivative of #G(t)# is

# G'(t)=lim_(h rarr 0) ( G(t+h)-G(t) ) / h #

So if # G(t) = (1-6t)/(5+t) # then;

# \ \ \ \ \ G(t+h) = (1-6(t+h))/(5+(t+h))#
# :. G(t+h) = (1-6t-6h)/(5+t+h)#

And so the derivative of #G((t)# is given by:

# \ \ \ \ \ G'(t) = lim_(h rarr 0) ( (1-6t-6h)/(5+t+h) - (1-6t)/(5+t) ) / h #
# :. G'(t) = lim_(h rarr 0) ( ((1-6t-6h)(5+t) - (1-6t)(5+t+h)) / ((5+t+h)(5+t)) ) / h #
# :. G'(t) = lim_(h rarr 0) ((5-30t-30h + t-6t^2-6ht) - (5+t+h -30t-6t^2-6th)) / (h(5+t+h)(5+t))#
# :. G'(t) = lim_(h rarr 0) (5-30t-30h + t-6t^2-6ht -5-t-h +30t+6t^2+6th) / (h(5+t+h)(5+t))#
# :. G'(t) = lim_(h rarr 0) (-30h -h ) / (h(5+t+h)(5+t))#
# :. G'(t) = lim_(h rarr 0) (-31h ) / (h(5+t+h)(5+t))#
# :. G'(t) = lim_(h rarr 0) (-31 ) / ((5+t+h)(5+t))#
# :. G'(t) = (-31 ) / ((5+t+0)(5+t))#
# :. G'(t) = (-31 ) / (5+t)^2#

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