$\dfrac{d}{d(x^n)}x^y = \dfrac{d}{d(x^n)}(x^n)^{\frac{y}{n}} = \dfrac{y}{n} (x^n) ^{\frac{y}{n} - 1} = \boxed{\dfrac{y}{n} (x^{y-n})}$ , where $y$ is a constant.

Solution 1

By the chain rule, we know that

$\displaystyle\begin{aligned} \frac{dx^y}{dx^n} \cdot \frac{dx^n}{dx} &= \frac{dx^y}{dx} \\ \frac{dx^y}{dx^n} \cdot n x^{n-1} &= y x^{y-1} \\ \frac{dx^y}{dx^n} &= \boxed{\frac{y}{n} x^{y-n}} \end{aligned}$

Solution 2

Let $a = x^n$ , then $x^y = a^{y/n}$ . Thus,

$\displaystyle\begin{aligned} \frac{dx^y}{dx^n} &= \frac{da^{y/n}}{da} \\ &= \frac{y}{n} a^{\frac{y}{n} - 1} \\ &= \frac{y}{n} (x^n)^{\frac{y}{n} - 1} \\ &= \frac{y}{n} x^{n \left( \frac{y}{n} - 1 \right)} \\ &= \boxed{\frac{y}{n} x^{y-n}} \end{aligned}$

Although this isn't the best way to go about solving the problem, we can do simple elimination of the options. We know that, for the case of y=n, the derivative is definitely going to be 1. That is one of the special conditions that the answer must fulfill. Since only one of the general solutions fulfills this condition, the answer is then obvious.

Again, not the best way to solve the problem but a smart method, especially in exam conditions. One of the ways in which you can find out, just by a glance, whether the equation is possibly correct is just by testing its validity in certain conditions and, in this scenario, only one of the equations passes the test.

Let u=x^n So x^y=(x^n)^(y/n)=u^(y/n) Hence d/du(u^(y/n))=y/n u^((y-n)/n) and substituting x back: d/d(x^n)=y/n x^(y-n).