How do you determine the derivative of this exponential function #f(x) = sqrt (3^x) / x^2#?

Question

Samantha Adkison · Accepted Answer

Write the function as #f(x)=\frac{3^{x/2}}{x^2}# and then use the Quotient Rule and the Chain Rule: #f'(x)=\frac{x^{2}\cdot ln(3)\cdot 3^{x/2}\cdot \frac{1}{2}-3^{x/2}\cdot 2x}{x^{4}}=\frac{3^{x/2}\cdot(ln(\sqrt{3})\cdot x-2)}{x^{3}}#

Isaiah Allen · Answer

undefined To find the derivative of the function $ f(x) = \frac{\sqrt{3^x}}{x^2} $, you can use the quotient rule followed by the chain rule. The quotient rule states that if you have a function of the form $ \frac{u(x)}{v(x)} $, the derivative is given by:

$$ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} $$

Now, let $ u(x) = \sqrt{3^x} $ and $ v(x) = x^2 $. We find the derivatives of $ u(x) $ and $ v(x) $ using the chain rule:

$$ u'(x) = \frac{d}{dx}(\sqrt{3^x}) = \frac{1}{2\sqrt{3^x}} \cdot \frac{d}{dx}(3^x) = \frac{1}{2\sqrt{3^x}} \cdot 3^x \cdot \ln(3) $$

$$ v'(x) = \frac{d}{dx}(x^2) = 2x $$

Now, applying the quotient rule:

$$ f'(x) = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{2\sqrt{3^x}} \cdot 3^x \cdot \ln(3) \cdot x^2 - \sqrt{3^x} \cdot 2x}{(x^2)^2} $$

$$ f'(x) = \frac{\frac{3^x \cdot \ln(3) \cdot x^2}{2\sqrt{3^x}} - 2x\sqrt{3^x}}{x^4} $$