How did addition come into stapel's answer on problem 5?
compared to:
\(\displaystyle y = \ln(3x^{2} + 5y^{2})\)
\(\displaystyle y' = \dfrac{1}{u}(6x + 10yy')\)
\(\displaystyle y' - 10yy' = \dfrac{1}{u}(6x)\)
\(\displaystyle y'[1 - 10y] = \dfrac{1}{u}(6x)\)
\(\displaystyle y' = \dfrac{1}{u}(6x)\dfrac{1}{(1 - 10y)}\)
\(\displaystyle y' = \dfrac{1}{3x^{2} + 5y^{2}}(6x)\dfrac{1}{(1 - 10y)}\)
\(\displaystyle y' = \dfrac{6x}{(3x^{2} + 5y^{2})(1 - 10y)}\)
\(\displaystyle \mbox{5. Find }\, y'\, \mbox{ if }\, y\, =\, \ln\left(3x^2\, +\, 5y^2\right)\)
Differentiate the log, and implicitly differentiate its argument.
. . . . .\(\displaystyle \frac{dy}{dx}\, =\, \left(\frac{1}{3x^2\, +\, 5y^2}\right)\left(6x\, +\, 10y\frac{dy}{dx}\right)\)
. . . . .\(\displaystyle \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}\, +\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}\)
. . . . .\(\displaystyle \frac{dy}{dx}\, -\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}\)
. . . . .\(\displaystyle \frac{dy}{dx}\left(1\, -\, \frac{10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}\)
. . . . .\(\displaystyle \frac{dy}{dx}\left(\frac{3x^2\, +\, 5y^2\, -\, 10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}\)
Divide through and simplify for the derivative. :wink:
compared to:
\(\displaystyle y = \ln(3x^{2} + 5y^{2})\)
\(\displaystyle y' = \dfrac{1}{u}(6x + 10yy')\)
\(\displaystyle y' - 10yy' = \dfrac{1}{u}(6x)\)
\(\displaystyle y'[1 - 10y] = \dfrac{1}{u}(6x)\)
\(\displaystyle y' = \dfrac{1}{u}(6x)\dfrac{1}{(1 - 10y)}\)
\(\displaystyle y' = \dfrac{1}{3x^{2} + 5y^{2}}(6x)\dfrac{1}{(1 - 10y)}\)
\(\displaystyle y' = \dfrac{6x}{(3x^{2} + 5y^{2})(1 - 10y)}\)
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