9514 1404 393
Answer:
a) y' = x^2(3x·ln(6x) +1)
b) y' = 6e^(3x)/(1 -e^(3x))^2
Step-by-step explanation:
The applicable rules for derivatives include ...
d(u^n)/dx = n·u^(n-1)·du/dx
d(uv)/dx = (du/dx)v +u(dv/dx)
d(e^u)/dx = e^u·du/dx
d(ln(u))/dx = 1/u·du/dx
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(a)
[tex]y=x^3\ln{(6x)}\\\\y'=3x^2\ln{(6x)}+\dfrac{x^3\cdot6}{6x}\\\\\boxed{\dfrac{dy}{dx}=3x^3\ln{(6x)}+x^2}[/tex]
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(b)
[tex]y=\dfrac{1+e^{3x}}{1-e^{3x}}=1+\dfrac{2}{1-e^{3x}}=1+2(1-e^{3x})^{-1}\\\\y'=-2(1-e^{3x})^{-2} (-3e^{3x})\\\\\boxed{\dfrac{dy}{dx}=\dfrac{6e^{3x}}{(1-e^{3x})^2}}[/tex]
