The graph of $\displaystyle y = 6(3x^2)^x$ has two horizontal tangent lines. Find the equations for both of them.

Alright, so my approach will be to find the derivative of the function using logarithmic differentiation or $\displaystyle b^x = e^\ln b^x$. Next, I'll set this derivative equal to zero and solve to obtain the tangent lines, since a tangent line occurs when $\displaystyle f'(x) = 0$. Unfortunately, both attempts have been unsuccessful.

**Logarithmic Differentiation**

$\displaystyle y = 6(3x^2)^x$

$\displaystyle \ln y = \ln [6(3x^2)^x]$

$\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} = \ln 6 + \ln (3x^2)^x$

$\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} = 0 + x\ln 3x^2$

$\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} = (1 \cdot \ln 3x^2) + x(\frac{1}{3x^2} \cdot 6x)$

$\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + \frac{6x^2}{3x^2}$

$\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} = \ln 3x^2 + 2$

$\displaystyle \frac{dy}{dx} = 6(3x^2)^x[\ln 3x^2 + 2]$

**Solve $\displaystyle y'$ For 0 To Find Horizontal Tangents Of Original Function**

$\displaystyle \ln(3x^2) + 2 = 0$

$\displaystyle \ln(3x^2) = -2 $

$\displaystyle e^{\ln(3x^2)} = e^{-2}$

$\displaystyle 3x^2 = e^{-2}$

$\displaystyle x^2 = \frac{e^{-2}}{3} $

$\displaystyle \sqrt{x^2} = \frac{\sqrt{e^{-2}}}{\sqrt{3}}$

$\displaystyle x = \pm \frac{1}{e\sqrt{3}}$

**Evaluate Original Function At $\displaystyle x$**

$\displaystyle y = \ln(3x^2)^x $

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{e\sqrt{3}})^2)^{\frac{1}{e\sqrt{3}}}$

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(3(\frac{1}{3e^2}))^{\frac{1}{e\sqrt{3}}}$

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(\frac{3}{3e^2})^{\frac{1}{e\sqrt{3}}}$

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(\frac{1}{e^2})^{\frac{1}{e\sqrt{3}}}$

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(e^{-2})^{\frac{1}{e\sqrt{3}}}$

$\displaystyle f(\frac{1}{e\sqrt{3}}) = 6(e^{\frac{-2}{1} \cdot \frac{\sqrt{3}}{3e}})$

$\displaystyle y = 6e^{\frac{-2\sqrt3}{3e}} $

(Substitute $\displaystyle -\frac{1}{e\sqrt{3}} $ to get the other solution)

The answer is said to be:

$\displaystyle y = 6e^{\frac{-2\sqrt3}{3e}} $ and $\displaystyle y = 6e^{\frac{2\sqrt3}{3e}} $

Which confirms that the solution is correct.

Thanks a lot for the help!