Find dy/dx by implicit differentiation. 3x²+3 = ln 5xy²
This is what i did: 6x = D (ln 5x + ln y²) But i was thinking i could also use this way:
6x = D(5xy²)/5xy²
Are both ways the same?
Differentiation by implicit
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$\begingroup$
algebra-precalculus
implicit-differentiation
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0@AhalleySamuel: You should add parenthesis to avoid ambiguity ... – 2017-02-25
1 Answers
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Yes, you are right.
If you try the first one, we get, $$\frac{\mathrm{d}}{\mathrm{d}x} (\ln 5x + \ln y^2) = \frac{1}{5x}(\frac{\mathrm{d}}{\mathrm{d}x}(5x)) = \frac{1}{x}$$
In case of the second, we get, $$\frac{\frac{\mathrm{d}}{\mathrm{d}x}(5xy^2)}{5xy^2} = \frac{5y^2}{5xy^2} = \frac{1}{x}$$ yielding the same answer as before.
Hope it helps.
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0Yea. Both ways arrive at the same results. Thanks man – 2017-03-04