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 October 5th, 2018, 06:20 AM #1 Newbie   Joined: Oct 2018 From: CA Posts: 1 Thanks: 0 Derivatives help Good morning, I was wondering if someone can give me a clear explanation of how to solve for derivatives? Please somebody help me... also what does it mean to solve for "the first derivative"? f(x)=⅓ x2 +7x - 4 f(x)= 5 x3+3x Thank you!
 October 5th, 2018, 06:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,713 Thanks: 1806 It presumably means that you should find the (first) derivative of each of the given functions of x by using the rules of differentiation, such as the power rule. Thanks from ProofOfALifetime
 October 5th, 2018, 06:59 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 Do you know what a first derivative is? There are two ways to find a first derivative. One is to work from the basic definition by taking the limit of the Newton quotient. This is usually done toward the start of a calculus course when they are trying to get you to understand what a derivative is. The other is to apply a set of general rules. First method. $f(x) = 3x^2 \implies \text { Newton quotient } = \dfrac{f(x + h) - f(x)}{h} =$ $\dfrac{3(x^2 + 2xh + h^2) - 3x^2}{h} = \dfrac{3x^2 + 6hx + 3h^2 - 3x^2}{h} = 6x + 3h^2.$ $\therefore \displaystyle \left ( \lim_{h \rightarrow 0} 6x + 3h \right ) = 6x = f'(x).$ Second method is to apply general rules. $\alpha (x) = k * \beta(x), \text { where } k \text { is a constant } \implies \alpha '(x) = k * \beta '(x).$ $\gamma (x) = x^n \implies \gamma '(x) = n * x^{(n-1)}.$ Put those two rules together and you get $f(x) = 3x^2 \implies f'(x) = 3 * ( 2 * x^{(2-1)}) = 6x.$ Thanks from ProofOfALifetime and Ljusbi

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