How do you use the definition of a derivative to find the derivative of #G(t) = (16t)/(5+t)#?
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To find the derivative of ( G(t) = \frac{16t}{5+t} ) using the definition of a derivative, we can follow these steps:

Start with the definition of the derivative: [ G'(t) = \lim_{h \to 0} \frac{G(t+h)  G(t)}{h} ]

Substitute the function ( G(t) ) into the definition: [ G'(t) = \lim_{h \to 0} \frac{\frac{16(t+h)}{5+(t+h)}  \frac{16t}{5+t}}{h} ]

Simplify the expression: [ G'(t) = \lim_{h \to 0} \frac{\frac{16t6h}{5+t+h}  \frac{16t}{5+t}}{h} ]

Find a common denominator: [ G'(t) = \lim_{h \to 0} \frac{(16t6h)(5+t)  (16t)(5+t+h)}{h(5+t)(5+t+h)} ]

Expand and simplify the numerator: [ G'(t) = \lim_{h \to 0} \frac{530t30h+t6th55t6t6h}{h(5+t)(5+t+h)} ] [ G'(t) = \lim_{h \to 0} \frac{30h6th}{h(5+t)(5+t+h)} ]

Cancel out the common factor of ( h ): [ G'(t) = \lim_{h \to 0} \frac{306t}{(5+t)(5+t+h)} ]

Take the limit as ( h ) approaches 0: [ G'(t) = \frac{306t}{(5+t)(5+t)} ]

Simplify the expression: [ G'(t) = \frac{306t}{(5+t)^2} ]
So, the derivative of ( G(t) ) with respect to ( t ) is ( \frac{306t}{(5+t)^2} ).
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