# How do you find the derivative of y = (1/3)^(x^2)?

Dec 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x \ln 3}{{3}^{{x}^{2}}}$

#### Explanation:

$y = {\left(\frac{1}{3}\right)}^{{x}^{2}}$

Take the natural logarithm of both sides.

$\ln y = \ln {\left(\frac{1}{3}\right)}^{{x}^{2}}$

Use laws of logarithms to simplify.

$\ln y = {x}^{2} \ln \left(\frac{1}{3}\right)$

Differentiate using implicit differentiation and the product rule.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x \left(\ln \left(\frac{1}{3}\right)\right) + {x}^{2} \left(0\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x \ln \left(\frac{1}{3}\right)}{\frac{1}{y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\frac{1}{3}\right)}^{{x}^{2}} 2 x \ln \left(\frac{1}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x {\left(\frac{1}{3}\right)}^{{x}^{2}} \ln 3 \to \text{since } \ln \left(\frac{1}{3}\right) = \ln \left({3}^{-} 1\right) = - \ln 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x \ln 3}{{3}^{{x}^{2}}}$

Hopefully this helps!