How do you use the definition of a derivative to find the derivative of $G (t) = \frac{4 t}{t + 1}$ $G(t)= (4t)/(t+1)$ ?

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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 $h$ .

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

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

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