calculus: Find y', if y = x^3 / 3^x (I get x^3=y'=3x^2...)
- Thread starter venialove
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pka
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Re: calculus Y PRIME
\(\displaystyle \begin{gathered} y = \frac{{x^3 }}{{3^x }} = \left( {x^3 } \right)\left( {3^{ - x} } \right) \hfill \\ y' = \left( {3x^2 } \right)\left( {3^{ - x} } \right) + \left( {x^3 } \right)\left[ { - 3^{ - x} \left( {\ln (3)} \right)} \right] \hfill \\ \end{gathered}\)venialove said:
Re: calculus Y PRIME
Hello, venialove!
Sorry, I have no idea what you did . . .
\(\displaystyle \text{Quotient Rule: }\;y' \;=\;\frac{3^x\!\cdot\!3x^2 - x^3\!\cdot\!3^x\!\cdot\!\ln3}{(3^x)^2}\)
\(\displaystyle \text{Factor: }\;y' \;=\;\frac{3^x\!\cdot\!x^2\left(3 - x\ln3)}{(3x)^2}\)
\(\displaystyle \text{Reduce: }\;y' \;=\;\frac{x^2(3 - x\ln3)}{3^x}\)
Hello, venialove!
Sorry, I have no idea what you did . . .
\(\displaystyle \text{Quotient Rule: }\;y' \;=\;\frac{3^x\!\cdot\!3x^2 - x^3\!\cdot\!3^x\!\cdot\!\ln3}{(3^x)^2}\)
\(\displaystyle \text{Factor: }\;y' \;=\;\frac{3^x\!\cdot\!x^2\left(3 - x\ln3)}{(3x)^2}\)
\(\displaystyle \text{Reduce: }\;y' \;=\;\frac{x^2(3 - x\ln3)}{3^x}\)
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Re: calculus Y PRIME
venialove said: