Ln Derivatives

How did addition come into stapel's answer on problem 5?
\(\displaystyle \mbox{5. Find }\, y'\, \mbox{ if }\, y\, =\, \ln\left(3x^2\, +\, 5y^2\right)\)

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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How did addition come into stapel's answer on problem 5?


compared to:

\(\displaystyle y = \ln(3x^{2} + 5y^{2})\)

\(\displaystyle y' = \dfrac{1}{u}(6x + 10yy')\)

\(\displaystyle y' - 10yy' = \dfrac{1}{u}(6x)\)

\(\displaystyle y'[1 - 10y] = \dfrac{1}{u}(6x)\)

\(\displaystyle y' = \dfrac{1}{u}(6x)\dfrac{1}{(1 - 10y)}\)

\(\displaystyle y' = \dfrac{1}{3x^{2} + 5y^{2}}(6x)\dfrac{1}{(1 - 10y)}\)

\(\displaystyle y' = \dfrac{6x}{(3x^{2} + 5y^{2})(1 - 10y)}\)

First, he multiplied 6x + 10yy' with 1/3x^2 + 5y^2 by means of distribution. You made a mistake by subtracting 10yy' while it is in a parenthesis with 6x. You needed to multiply the entire thing with the outside fraction first. So, yours should have been 6x/u + 10yy'/ u.

I am pretty pretty sure that's where you went wrong. :)
 
No, this is what I put. Sorry, I stayed up all night doing other homework so I was sorta half as sleep when I typed. :(

View attachment 3341

By any chance do you have what the computer says is the "correct" answer? I ask for that because what you state above is correct. Sometimes the computer's answer is simplified in a way that they consider your answer incorrect when it really isn't.
 
First, he multiplied 6x + 10yy' with 1/3x^2 + 5y^2 by means of distribution. You made a mistake by subtracting 10yy' while it is in a parenthesis with 6x. You needed to multiply the entire thing with the outside fraction first. So, yours should have been 6x/u + 10yy'/ u.

I am pretty pretty sure that's where you went wrong. :)

Ok, i see the error. Thanks
 
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