For the first derivative I got -x/y, is that correct? According to the book the second derivative is -2x/y^5. How does this happen? When I take the derivative of -x/y, I get (y[-1]-[y`]-x)/y^2 (quotient rule). To get the derivative of the original equation I did chain rule and got 3x^2+ (y`)3y^2 =>y`= -3x^2/-3y^2 => -x/y
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Derivative and 2nd derivative of x^3+y^3=1?
- Thread starter pr1nny
- Start date
Differentiating the given function implicitly gives 3x^2 + 3y^2y' = 0 and, therefore, y' = -x^2 / y^2. Differentiating y' gives
y'' = [-2xy^2 + 2x^2yy'] / y^3 which simplifies to -2x/y^5 [x^3 + y^3]. But we are given x^3 + y^3 = 1 and hence y'' = -2x/y^5.
Regards,
Rich B.
y'' = [-2xy^2 + 2x^2yy'] / y^3 which simplifies to -2x/y^5 [x^3 + y^3]. But we are given x^3 + y^3 = 1 and hence y'' = -2x/y^5.
Regards,
Rich B.
BigGlenntheHeavy
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\(\displaystyle First \ one, \ assuming \ we \ want \ y' \ = \ \frac{dy}{dx} \ of \ x^3+y^3 \ = \ 1.\)
\(\displaystyle Implicit \ differentation: \ 3x^2+3y^2y' \ = \ 0, \ \implies \ y' \ = \ \frac{-x^2}{y^2}\)
\(\displaystyle Explicit \ differentation: \ y \ = \ (1-x^3)^{1/3}, \ y' \ = \ \frac{-x^2}{(1-x^3)^{2/3}}\)
\(\displaystyle I'll \ leave \ it \ up \ to \ you \ to \ show \ that \ y^2 \ = \ (1-x^3)^{2/3}.\)
\(\displaystyle Implicit \ differentation: \ 3x^2+3y^2y' \ = \ 0, \ \implies \ y' \ = \ \frac{-x^2}{y^2}\)
\(\displaystyle Explicit \ differentation: \ y \ = \ (1-x^3)^{1/3}, \ y' \ = \ \frac{-x^2}{(1-x^3)^{2/3}}\)
\(\displaystyle I'll \ leave \ it \ up \ to \ you \ to \ show \ that \ y^2 \ = \ (1-x^3)^{2/3}.\)
nikkor180 & edit said:y' = -x^2 / y^2. Differentiating y' gives
\(\displaystyle >>\) y'' = [-2xy^2 + 2x^2yy'] / y^3 \(\displaystyle <<\)
which simplifies to -2x/y^5 [x^3 + y^3].
But we are given x^3 + y^3 = 1 and hence y'' = -2x/y^5.
nikkor180,
the highlighted expression needs a different exponent on the y variable in the denominator:
y'' = [-2xy^2 + 2(x^2)yy']/(y^2)^2 =
[-2xy^2 + 2(x^2)yy']/y^4 =
\(\displaystyle \frac{-2xy^2 + 2x^2yy'}{y^4} =\)
\(\displaystyle \frac{y(-2xy^2 + 2x^2yy')}{y(y^4)} =\)
\(\displaystyle \frac{-2xy^3 + 2x^2y^2y'}{y^5} =\)
Substitute in the y' expression:
\(\displaystyle \frac{-2xy^3 + 2x^2y^2(\frac{-x^2}{y^2})}{y^5}=\)
\(\displaystyle \frac{-2xy^3 - 2x^4}{y^5} =\)
\(\displaystyle \frac{-2x}{y^5}(y^3 + x^3) =\)
\(\displaystyle Note: \ \ x^3 + y^3 = 1 \ (or \ y^3 + x^3 = 1) \ from \ the \ given.\)
\(\displaystyle \frac{-2x}{y^5}(1) =\)
\(\displaystyle \frac{-2x}{y^5}\)