How do you use the definition of a derivative to find the derivative of #G(t)= (4t)/(t+1)#?
The limit definition of a derivative states that
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To find the derivative of ( G(t) = \frac{4t}{t+1} ) using the definition of a derivative, follow these steps:

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

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} ]

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)} ]

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)} ]

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

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

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

Simplify the expression: [ G'(t) = \frac{4}{(t+1)^2} ]
Therefore, the derivative of ( G(t) ) is ( \frac{4}{(t+1)^2} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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