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

Question

Emma Aldridge · Accepted Answer

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

Isabella Ackerman · Answer

undefined To find the derivative of $ G(t) = \frac{1-6t}{5+t} $ using the definition of a derivative, we can follow these steps:

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

2. Substitute the function $ G(t) $ into the definition:
   $$ G'(t) = \lim_{h \to 0} \frac{\frac{1-6(t+h)}{5+(t+h)} - \frac{1-6t}{5+t}}{h} $$

3. Simplify the expression:
   $$ G'(t) = \lim_{h \to 0} \frac{\frac{1-6t-6h}{5+t+h} - \frac{1-6t}{5+t}}{h} $$

4. Find a common denominator:
   $$ G'(t) = \lim_{h \to 0} \frac{(1-6t-6h)(5+t) - (1-6t)(5+t+h)}{h(5+t)(5+t+h)} $$

5. Expand and simplify the numerator:
   $$ G'(t) = \lim_{h \to 0} \frac{5-30t-30h+t-6th-5-5t-6t-6h}{h(5+t)(5+t+h)} $$
   $$ G'(t) = \lim_{h \to 0} \frac{-30h-6th}{h(5+t)(5+t+h)} $$

6. Cancel out the common factor of $ h $:
   $$ G'(t) = \lim_{h \to 0} \frac{-30-6t}{(5+t)(5+t+h)} $$

7. Take the limit as $ h $ approaches 0:
   $$ G'(t) = \frac{-30-6t}{(5+t)(5+t)} $$

8. Simplify the expression:
   $$ G'(t) = \frac{-30-6t}{(5+t)^2} $$

So, the derivative of $ G(t) $ with respect to $ t $ is $ \frac{-30-6t}{(5+t)^2} $.