August 26th, 2017, 12:14 AM  #1 
Newbie Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0  Differential Calculus
How to differentiate these functions using the definition ug derivatives? ½t⁴  5t 3 (x² 2)² 
August 26th, 2017, 05:27 AM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 1,857 Thanks: 964  Quote:
$\dfrac{d}{dt}\left(\dfrac 1 2 t^4  5t  3\right) =$ $\displaystyle \lim_{h>0} \dfrac{\left(\frac 1 2 (t+h)^4  5(t+h)  3\right)  \left(\frac 1 2 t^4  5t  3\right)}{h} = $ $\displaystyle \lim_{h>0} \dfrac{\frac{h^4}{2}+2 h^3 t+3 h^2 t^2+h \left(2 t^35\right)}{h} = $ $\displaystyle \lim_{h>0} \frac{h^3}{2}+2h^2 t + 3h t^2 + (2t^35)=$ $2t^3  5$ I leave the second one to you.  
August 26th, 2017, 05:55 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 997 Thanks: 410 
The definition involves a limit and a quotient, namely, $f'(x) \equiv \displaystyle \lim_{h \rightarrow 0} \dfrac{f(x + h)  f(x)}{h}.$ My recommendation is to simplify the quotient before even thinking about limits. That means in practice simplifying the denominator and then dividing. Just algebra. First problem $f(t + h)  f(t) = 0.5(t + h)^4  5(t + h)  3  (0.5t^4  5t  3) =$ $0.5(t^4 + 4ht^3 + 6h^2t^2 + 4h^3t + h^4)  5t  5h  3  0.5t^4 + 5t + 3 =$ $2ht^3 + 3h^2t^2 + 2h^3t + 0.5h^4  5h.$ $\therefore \dfrac{f(t + h)  f(t)}{h} = \dfrac{2ht^3 + 3h^2t^2 + 2h^3t + 0.5h^4  5h}{h} =$ $2t^3 + 3ht^2 + 2h^2t + 0.5h^3  5 = 2t^3  5 + h(3t^2 + 2ht + 0.5h^2).$ $\displaystyle f'(t) = \lim_{h \rightarrow 0}\dfrac{f(t + h)  f(t)}{h} = \lim_{h \rightarrow 0}(2t^3  5 + h(3t^2 + 2ht + 0.5h^2)) = WHAT$ EDIT: There is no difference in the result from doing everything at once as romsek does and leaving limits to the end. I just think ignoring limits until the end may be easier for students starting out. Logically the two methods are exactly equivalent. Last edited by JeffM1; August 26th, 2017 at 06:00 AM. 

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calculus, differential 
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