ursoweird123
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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.
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Calculus equation using delta
- Thread starter ursoweird123
- Start date
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.
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
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle g(x) \ = \ x^2+3x+1\)
\(\displaystyle g'(x) \ = \ \lim_{\Delta x\to0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\)
\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)
\(\displaystyle g'(x) \ = \ \lim_{\Delta x\to0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\)
\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)
Hello, ursoweird123!
This asks for the Difference Quotient.
. . Why is everyone deriving the derivative ?
\(\displaystyle \text{There are three steps to the Difference Quotient: }\)
. . \(\displaystyle \text{[1] Find }g(x+\Delta x)\:\text{ . . . Replace }x\text{ with }x+\Delta x,\:\text{ and simplify.}\)
. . \(\displaystyle \text{[2] Subtract }g(x)\!:\;\text{ . . . Subtract the original function.}\:\text{ and simplify.}\)
. . \(\displaystyle \text{[3] Divide by }\Delta x\:\text{ . . . and simplify.}\)
\(\displaystyle \text{Here we go!}\)
\(\displaystyle [1]\;\;g(x+\Delta x) \;=\;(x+\Delta x)^2 + 3(x+\Delta x) + 1\)
. . . . . . . . . . \(\displaystyle =\;x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1\)
\(\displaystyle [2]\;\;g(x+\Delta x) - g(x) \;=\;\bigg[x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1\bigg] - \bigg[x^2 + 3x + 1\bigg]\)
. . . . . . . . . . . . . .\(\displaystyle =\;2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)\)
\(\displaystyle [3]\;\;\frac{g(x+\Delta x) - g(x)}{\Delta x} \;=\; \frac{2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)}{\Delta x}\)
. . . . . . . . . . . . . . . \(\displaystyle =\;\frac{(\Delta x)\bigg[2x + (\Delta x) + 3\bigg]}{\Delta x}\)
. . . . . . . . . . . . . . . \(\displaystyle =\;2x + (\Delta x) + 3\)
This asks for the Difference Quotient.
. . Why is everyone deriving the derivative ?
\(\displaystyle \text{Find: }\;\frac{g(x+\Delta x) - g(x)}{\Delta x}\;\text{ for }g(x)\,=\,x^2+3x-1\)
\(\displaystyle \text{There are three steps to the Difference Quotient: }\)
. . \(\displaystyle \text{[1] Find }g(x+\Delta x)\:\text{ . . . Replace }x\text{ with }x+\Delta x,\:\text{ and simplify.}\)
. . \(\displaystyle \text{[2] Subtract }g(x)\!:\;\text{ . . . Subtract the original function.}\:\text{ and simplify.}\)
. . \(\displaystyle \text{[3] Divide by }\Delta x\:\text{ . . . and simplify.}\)
\(\displaystyle \text{Here we go!}\)
\(\displaystyle [1]\;\;g(x+\Delta x) \;=\;(x+\Delta x)^2 + 3(x+\Delta x) + 1\)
. . . . . . . . . . \(\displaystyle =\;x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1\)
\(\displaystyle [2]\;\;g(x+\Delta x) - g(x) \;=\;\bigg[x^2 + 2x(\Delta x) + (\Delta x)^2 + 3x + 3(\Delta x) + 1\bigg] - \bigg[x^2 + 3x + 1\bigg]\)
. . . . . . . . . . . . . .\(\displaystyle =\;2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)\)
\(\displaystyle [3]\;\;\frac{g(x+\Delta x) - g(x)}{\Delta x} \;=\; \frac{2x(\Delta x) + (\Delta x)^2 + 3(\Delta x)}{\Delta x}\)
. . . . . . . . . . . . . . . \(\displaystyle =\;\frac{(\Delta x)\bigg[2x + (\Delta x) + 3\bigg]}{\Delta x}\)
. . . . . . . . . . . . . . . \(\displaystyle =\;2x + (\Delta x) + 3\)