Find y' if y = x^3/3^x
well there are 5 ways to solve this problem
one of them is by using the quotient rule, or the product rule
\[y =\frac{ x^{3} }{ 3^{x} }\]
this can be rewritten as
\[y = x^{3}. 3^{-x}\]
you know that the product rule is udv +vdu
\[\frac{ dy }{ dx } = 3^{-x}3x^{2} + x^{3}\ln(3)(-3^{-x})\]
\[\frac{ dy }{ dx } = 3^{-x}3x^{2} - x^{3}\ln(3)3^{-x}\]
\[\frac{ dy }{ dx } = x^{2}3^{-x}[3 - x\ln(3)]\]
3^(-x) , not 3^x
use wolfram to get 3*x^2/3^x-x^3*ln(3)/3^x
\[3\,{\frac {{x}^{2}}{{3}^{x}}}-{\frac {{x}^{3}\ln \left( 3 \right) }{{ 3}^{x}}}\]
well when you take 3^(x) from denominator to numerator it becomes 3^(-x)
and timo used a software and i can see his result is similar to mine
to solve 3^(-x) use d operator differential method
if anyone requires a proof let me know
i will solve ur partial derivative it is simple just give me a moment i am on the phone mr Hoa
don't be upset. we try to help each other, right? if i see your mistake, i just help you to correct it. i, myself need help, too. Sorry if i bother you.
i am not bothered my friend
my answer is correct
100% agree with you that we must solve the problem by hand, because in test, we don't have computer to search the result
\[y = \frac{ x^3 }{ 3^{x} }\]
now what happens if you take 3^x to the numerator
you ask me or the asker?
\[\frac{ 3^{-x} }{ 1 }\]
why? I don't understand, please, explain
well i wanted to simplify the problem by using the product rule and avoiding the quiescent rule
so i took the 3^(x) from the denominator to the numerator so it became 3^(-x)
using basic log and anti log rules
yeah, you are right, because the product must be x^3 * 3^(-x) , your typing lack of (-) . That's it. the whole thing is right
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