\[\begin{align*} \boxed{\dfrac{d}{dx} \left(\dfrac3x\right) = -\dfrac3{x^2}} \end{align*}\]
Solving for the Derivative
We may pull the constant, $3$, out of the derivative and focus on $\dfrac{d}{dx} \left(\dfrac1x\right)$. This is the same as $x^{-1}$, so we may apply the power rule:
\[\begin{align*} \dfrac{d}{dx} x^n = nx^{n-1} \end{align*}\]
Doing this with $x^{-1}$, we get our derivative is equal to
\[\begin{align*} \dfrac{d}{dx} \left(\dfrac3x\right) &= 3 \dfrac{d}{dx} \left(\dfrac1x\right)\\ &=3 \dfrac{d}{dx} x^{-1}\\ &=3 \cdot -1 \cdot x^{-2}\\ &= -3 \cdot \dfrac{1}{x^2}\\ &=\boxed{-\dfrac{3}{x^2}} \end{align*}\]
The key to this problem is rewriting it in a way that the power rule handles the heavy lifting. It is an important skill to know how to rewrite the problem as a negative exponent when $x$ is in the denominator. We always want positive exponents as the final answer so make sure to reconvert into a fraction at the end!
