Tangent Equation for Diff Problem

[Jason76]

Find tangent equation for point (1,0)

\(\displaystyle y = \dfrac{x^{2} - 1}{x^{2} + x + 1}\)

\(\displaystyle y' = \dfrac{[x^{2} - 1][2x] -[x^{2} - 1][2x + 1]}{(x^{2} + x + 1)^{2}}\)

\(\displaystyle y' = \dfrac{2x^{7} + 2x^{2} + 2x - 2x^{3} + x^{2} - 2x - 1}{(x^{2} + x + 1)^{2}}\)

\(\displaystyle y' = \dfrac{2x^{7} + 2x^{2} - 2x^{3} + x^{2} - 1}{(x^{2} + x + 1)^{2}}\)

\(\displaystyle y' = \dfrac{2(x^{7} + x^{2} - x^{3}) + x^{2} - 1}{(x^{2} + x + 1)^{2}}\)

 

[stapel]

Find tangent equation for point (1,0)

\(\displaystyle y = \dfrac{x^{2} - 1}{x^{2} + x + 1}\)

\(\displaystyle y' = \dfrac{[x^{2} - 1][2x] -[x^{2} - 1][2x + 1]}{(x^{2} + x + 1)^{2}}\)

How did you get the last line above? It looks like you're saying that the derivative of \(\displaystyle x^2\, +\, x\, +\, 1\) is just \(\displaystyle 2x\)...?

 

[Jason76]

How did you get the last line above? It looks like you're saying that the derivative of \(\displaystyle x^2\, +\, x\, +\, 1\) is just \(\displaystyle 2x\)...?

I'm using \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

 

[Deleted member 4993]

I'm using \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

you have

f = x²- 1 → f' = 2x

g = x²+ x + 1 → g' = 2x +1

Now check and re-check your expression for \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\)

 

[Deleted member 4993]

Find tangent equation for point (1,0)

\(\displaystyle y = \dfrac{x^{2} - 1}{x^{2} + x + 1}\)

\(\displaystyle y' = \dfrac{[x^{2} - 1][2x] -[x^{2} - 1][2x + 1]}{(x^{2} + x + 1)^{2}}\) ....................... Incorrect

\(\displaystyle y' = \dfrac{2x^{7} + 2x^{2} + 2x - 2x^{3} + x^{2} - 2x - 1}{(x^{2} + x + 1)^{2}}\) ........... where did x⁷ come from ??!!

\(\displaystyle y' = \dfrac{2x^{7} + 2x^{2} - 2x^{3} + x^{2} - 1}{(x^{2} + x + 1)^{2}}\)

\(\displaystyle y' = \dfrac{2(x^{7} + x^{2} - x^{3}) + x^{2} - 1}{(x^{2} + x + 1)^{2}}\)

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