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: 20,968 Thanks: 2217 
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.

October 5th, 2018, 06:59 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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. $\alpha (x) = k * \beta(x), \text { where } k \text { is a constant } \implies \alpha '(x) = k * \beta '(x).$ $\gamma (x) = x^n \implies \gamma '(x) = n * x^{(n1)}.$ Put those two rules together and you get $f(x) = 3x^2 \implies f'(x) = 3 * ( 2 * x^{(21)}) = 6x.$ 

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