I came across some derivatives w.r.t functions such as, $\frac{d(7^x)}{d(x^7)}$,
I tried plotting their graphs and seeing how we can relate the change in the value of $7^x$ as the function $x^7$ is changing. Can someone please provide visualization for this specific example or just a general intuitive explanation?
Thank you!
In general, $\dfrac{df(g(x))}{dg(x)}$ means $\dfrac{df(u)}{du}\Biggr|_{\large{u=g(x)}}=f'(g(x)) \, .$ You want to consider the behaviour of the function $f$ when $g(x)$ is incremented by an infinitely small amount $dg(x)$ (notice that the fact that $g$ itself is a function is irrelevant here—you simply want to treat $g(x)$ as a variable like any other). So informally, we have $$ \frac{df(g(x))}{dg(x)}=\frac{f(g(x)+dg(x))-f(g(x))}{dg(x)} \, . $$ Nowadays, we tend to think in terms of limits instead of "infinitesimals", and so we would write $$ \dfrac{df(g(x))}{dg(x)}=\lim_{h \to 0}\frac{f\left(g(x)+h\right)-f(g(x))}{h} \, . $$ In your example, $$ \frac{d(7^x)}{d(x^7)}=\frac{d\left(7^{\sqrt[7]{x^7}}\right)}{d(x^7)}=\frac{d\left(7^{\sqrt[7]{u}}\right)}{du}\Biggr|_{\large{u=x^7}}=\lim_{h \to 0}\frac{7^{\sqrt[7]{u+h}}-7^{\sqrt[7]{u}}}{h}\Biggr|_{\large{u=x^7}}=\lim_{h \to 0}\frac{7^{\sqrt[7]{x^7+h}}-7^{\sqrt[7]{x^7}}}{h} \, . $$ This tells us that the quantities $x^7$ and $7^x$ are related by the function $u\mapsto 7^{\sqrt[7]{u}}$. Here is a plot of $7^{\sqrt[7]{u}}$ against $u$:
Using the chain rule multiple times tells us that \begin{align} \frac{d\left(7^{\sqrt[7]{u}}\right)}{du}\Biggr|_{\large{u=x^7}} &= \frac{1}{7} \cdot \log 7 \cdot 7^{\sqrt[7]{u}} \cdot u^{-6/7} \\[5pt] &= \frac{7^x \cdot \log 7}{7x^6} \, . \end{align} Note that this result agrees with writing $$ \frac{d(7^x)}{d(x^7)} = \frac{d(7^x)}{dx} \cdot \frac{dx}{d(x^7)} \, , $$ and using the inverse function theorem, which is probably the quickest way to arrive at the answer.