How do you use the definition of a derivative to find the derivative of #G(t)= (4t)/(t+1)#?

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Answer 1 · 2016-01-24T19:03:32

The limit definition of a derivative states that

#G'(t)=lim_(hrarr0)(G(t+h)-G(t))/h#

Since #G(t)=(4t)/(t+1)#, we know that #G(t+h)=(4t+4h)/(t+h+1)#. Thus,

#G'(t)=lim_(hrarr0)((4t+4h)/(t+h+1)-(4t)/(t+1))/h#

Multiply the numerator and denominator by #(t+h+1)(t+1)# to clear the fractions.

#=lim_(hrarr0)((4t+4h)/(t+h+1)-(4t)/(t+1))/h*((t+h+1)(t+1))/((t+h+1)(t+1))#

#=lim_(hrarr0)((4t+4h)(t+1)-4t(t+h+1))/(h(t+h+1)(t+1))#

Distribute.

#=lim_(hrarr0)(4t^2+4t+4ht+4h-4t^2-4ht-4t)/(h(t+h+1)(t+1))#

#=lim_(hrarr0)(4h)/(h(t+h+1)(t+1))#

#=lim_(hrarr0)4/((t+h+1)(t+1))#

Now we can evaluate the limit by plugging in #0# for #h#.

#=4/((t+1)(t+1))#

#G'(t)=4/(t+1)^2#