Derivatives

[Johnmoon]

Can't seem to find the answers for this no matter how many videos I watch. Please help.

 

[JeffM]

[MATH]\text {Given: } f(x) = 6x + 11.[/MATH]
[MATH]\text {Thus } f(x + h) = 6(x + h) + 11 = 6x + 6h + 11 \implies[/MATH]
[MATH]f(x + h) - f(x) = (6x + 6h + 11) - (6x + 11) = 6x - 6x + 11 - 11 + 6h = 6h.[/MATH]
[MATH]\therefore \dfrac{f(x + h) - f(x)}{h} = \dfrac{6h}{h} = WHAT?[/MATH]
My recommendation is initially to proceed step by step. So first calculate f(x + h). Then calculate the difference
f(x + h) - f(x). Then divide by h. Now simplify. In other words, calculate the Newton quotient step by step without any worrying about limits. Only when you have the Newton quotient in its simplest form move on to taking the limit.

Does this help?

Now give a try to calculating the derivative of

[MATH]f(x) = 6x^2 + 11.[/MATH]

 

[pka]

Can't seem to find the answers for this no matter how many videos I watch. Please help.View attachment 12598

\(\displaystyle \begin{align*}\frac{f(x+h-f(x)}{h}&=\frac{[6(x+h)+1]-[6x+1]}{h} \\&=\frac{6h}{h}\\&=6 \end{align*}\)
Thus \(\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) = f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} 6 = 6\)

 

[Dr.Peterson]

Can't seem to find the answers for this no matter how many videos I watch. Please help.View attachment 12598

You've been shown the answer; but I suspect that where you typed (6(x+h)+11-(6x+11))/h twice, they intend for you to type the numerator and denominator separately:

6(x+h)+11-(6x+11)​

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