Calculus equation using delta

[ursoweird123]

Find (g(x+[delta]x)-g(x))/[delta]x for g(x)=x^2+3x-1. I really have no idea how to do this because we never learned it. This is summer work so I don't have anyone to ask for help. Thanks.

 

[galactus]

This appears to be the defintion of a derivative. What is also called 'first principles'.

I am going to use h instead of delta x and f instead of g. They are more common than the notation you presented.

Same thing though.

$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Find the derivative of $\displaystyle f(x)=x^{2}+3x-1$ using first principles or the definition of a derivative. If done correctly, we should end up with the derivative of $\displaystyle x^{2}+3x-1$. Which is 2x+3.

The algebra is mostly the booger in these.

$\displaystyle \lim_{h\to 0}\frac{\overbrace{(x+h)^{2}+3(x+h)-1}^{\text{f(x+h)}}-\overbrace{(x^{2}+3x-1)}^{\text{f(x)}}}{h}$

See?. For the f(x+h), sub in x+h in for x in the given expression.

Expand out the numerator:

$\displaystyle \lim_{x\to h}\frac{2xh+h^{2}+3h}{h}$

Now, this is the same as $\displaystyle \lim_{h\to 0}\frac{2x\not{h}}{\not{h}}+\lim_{h\to 0}\frac{h^{\not{2}^{1}}}{\not{h}}+\lim_{h\to 0}\frac{3\not{h}}{\not{h}}$

Note how the h's cancel and we are left with $\displaystyle 2x+h+3$.

Per the limit, as h goes to 0, the center term h=0 and we have 2x+3 as required.

The algebra here was not that bad, but some can be a dilly.

 

[BigGlenntheHeavy]

$\displaystyle g(x) \ = \ x^2+3x+1$

$\displaystyle g'(x) \ = \ \lim_{\Delta x\to0}\frac{g(x+\Delta x)-g(x)}{\Delta x}$

$\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.$

 

[soroban]

Hello, ursoweird123!

This asks for the Difference Quotient.

. . Why is everyone deriving the derivative ?

$\displaystyle \text{Find: }\;\frac{g(x+\Delta x) - g(x)}{\Delta x}\;\text{ for }g(x)\,=\,x^2+3x-1$

$\displaystyle \text{There are three steps to the Difference Quotient: }$

. . $\displaystyle \text{[1] Find }g(x+\Delta x)\:\text{ . . . Replace }x\text{ with }x+\Delta x,\:\text{ and simplify.}$

. . $\displaystyle \text{[2] Subtract }g(x)\!:\;\text{ . . . Subtract the original function.}\:\text{ and simplify.}$

. . $\displaystyle \text{[3] Divide by }\Delta x\:\text{ . . . and simplify.}$


$\displaystyle \text{Here we go!}$


$\displaystyle [1]\;\;g(x+\Delta x) \;=\;(x+\Delta x)^2 + 3(x+\Delta x) + 1$

. . . . . . . . . . $\displaystyle =\;x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1$


$\displaystyle [2]\;\;g(x+\Delta x) - g(x) \;=\;\bigg[x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1\bigg] - \bigg[x^2 + 3x + 1\bigg]$

. . . . . . . . . . . . . .$\displaystyle =\;2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)$


$\displaystyle [3]\;\;\frac{g(x+\Delta x) - g(x)}{\Delta x} \;=\; \frac{2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)}{\Delta x}$

. . . . . . . . . . . . . . . $\displaystyle =\;\frac{(\Delta x)\bigg[2x + (\Delta x) + 3\bigg]}{\Delta x}$

. . . . . . . . . . . . . . . $\displaystyle =\;2x + (\Delta x) + 3$