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 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
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Quote:
 Originally Posted by Grntxavier How to differentiate these functions using the definition ug derivatives? ½t⁴ - 5t- 3 (x² -2)²
$\dfrac{d}{dt}f(t) = \displaystyle \lim_{h->0} \dfrac{f(t+h)-f(t)}{h}$

$\dfrac{d}{dt}\left(\dfrac 1 2 t^4 - 5t - 3\right) =$

$\displaystyle \lim_{h->0} \dfrac{\left(\frac 1 2 (t+h)^4 - 5(t+h) - 3\right) - \left(\frac 1 2 t^4 - 5t - 3\right)}{h} =$

$\displaystyle \lim_{h->0} \dfrac{\frac{h^4}{2}+2 h^3 t+3 h^2 t^2+h \left(2 t^3-5\right)}{h} =$

$\displaystyle \lim_{h->0} \frac{h^3}{2}+2h^2 t + 3h t^2 + (2t^3-5)=$

$2t^3 - 5$

I leave the second one to you.

 August 26th, 2017, 05:55 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 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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