Derivative x^3e^(-1/x) +3^xln(1/x)

[Sophie]

Question: Find the derivative in each case. Simplify your answer as far as possible>

\(\displaystyle \L\\\begin{array}{l}
g(x) = x^3 e^{ - 1/x} + 3^x \ln \left( {\frac{1}{x}} \right) \\
\\
g(x) = x^3 e^{ - 1/x} + 3^x \ln \left( {\frac{1}{x}} \right) \\
\\
g'(x) = 3x^2 e^{ - 1/x} + x^3 e^{ - 1/x} \frac{1}{{x^2 }} + 3^x \ln 3\ln \left( {\frac{1}{x}} \right) + 3^x x \\
\\
g'(x) = xe^{ - 1/x} \left( {3x + 1} \right) + 3^x \left( {\ln 3\ln \left( {\frac{1}{x}} \right) + x} \right) \\
\end{array}\)

I found the derivative using ln as well, but the solutions were different, if someone could check this I would be greatful.

Thanks Sophie

 

[soroban]

Hello, Sophie!

Some advance simplifying would have helped . . .

Find the derivative in each case; simplify.

. . \(\displaystyle \L g(x) \:= \:x^3\cdot e^{(-1/x)}\, +\, 3^x\cdot\ln\left(\frac{1}{x}\right)\)

We have: \(\displaystyle \L\:g(x) \;=\;x^3\cdot e^{(-1/x)}\,+\,3^x\cdot\ln\left(x^{^{-1}}\right)\)

. . . . . . . . \(\displaystyle \L g(x)\;=\;x^3\cdot e^{(-x^{^{-1}})} \,-\,3^x\cdot\ln(x)\)


Then: \(\displaystyle \L\:g'(x)\;=\;3x^2\cdot e^{^{1/x}}\,+\,x^3\cdot e^{^{1/x}}\cdot\left(x^{-2}\right)\,-\,3^x\cdot\ln(3)\cdot\ln(x) \,-\,3^x\left(\frac{1}{x}\right)\)

. . . . . \(\displaystyle \L g'(x)\;=\;3x^2\cdot e^{^{1/x}}\,+\,x\cdot e^{^{1/x}}\,-\,3^x\cdot\ln(3)\cdot\ln(x)\,-\,\frac{3^x}{x}\)

. . . . . \(\displaystyle \L g'(x)\;=\;xe^{^{1/x}}(3x\,+\,1)\,-\,3^x\left[\ln(3)\cdot\ln(x)\,+\,\frac{1}{x}\right]\)

 

[Sophie]

Thanks