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

Question

Elijah Abram · Accepted Answer

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#

Charles Arocha · Answer

undefined To find the derivative of $ G(t) = \frac{4t}{t+1} $ using the definition of a derivative, follow these steps:

1. Begin with the definition of the derivative:
$$ G'(t) = \lim_{{h \to 0}} \frac{G(t+h) - G(t)}{h} $$

2. Substitute $ G(t) = \frac{4t}{t+1} $ into the definition:
$$ G'(t) = \lim_{{h \to 0}} \frac{\frac{4(t+h)}{t+h+1} - \frac{4t}{t+1}}{h} $$

3. Simplify the expression inside the limit:
$$ G'(t) = \lim_{{h \to 0}} \frac{(4(t+h)(t+1) - 4t(t+h+1))}{h(t+1)(t+h+1)} $$

4. Expand and simplify the numerator:
$$ G'(t) = \lim_{{h \to 0}} \frac{4t + 4h + 4t + 4 - 4t - 4ht - 4t - 4h}{h(t+1)(t+h+1)} $$

5. Further simplify the expression:
$$ G'(t) = \lim_{{h \to 0}} \frac{4h}{h(t+1)(t+h+1)} $$

6. Cancel out common terms:
$$ G'(t) = \lim_{{h \to 0}} \frac{4}{(t+1)(t+h+1)} $$

7. Substitute $ h = 0 $ into the expression:
$$ G'(t) = \frac{4}{(t+1)(t+1)} $$

8. Simplify the expression:
$$ G'(t) = \frac{4}{(t+1)^2} $$

Therefore, the derivative of $ G(t) $ is $ \frac{4}{(t+1)^2} $.