Answer: Proof below; \(\dfrac{d(7^x)}{d(x^7)} = \dfrac{7^x\log 7}{7x^6}\).
Proof: Let \(\delta t\) be a small increment in \(t\), and \(\delta x\), \(\delta y\) the corresponding increments in \(x\) and \(y\). Since \(\dfrac{dx}{dt} \ne 0\), \(\delta x \ne 0\) for small \(\delta t \ne 0\). Then
\(\dfrac{\delta y}{\delta x} = \dfrac{\delta y/\delta t}{\delta x/\delta t}\).
As \(x = f(t)\) is differentiable, it is continuous, so \(\delta x \to 0\) as \(\delta t \to 0\). Taking limits as \(\delta t \to 0\):
\(\displaystyle\lim_{\delta x \to 0}\dfrac{\delta y}{\delta x} = \dfrac{\lim_{\delta t \to 0}\frac{\delta y}{\delta t}}{\lim_{\delta t \to 0}\frac{\delta x}{\delta t}}\), i.e. \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\), since \(y\) is a differentiable function of \(x\) and \(\dfrac{dx}{dt} \ne 0\).
Hence: Let \(u = 7^x\) and \(v = x^7\), both functions of \(x\) (here \(x\) plays the role of the parameter).
\(\dfrac{du}{dx} = 7^x\log 7\), \(\dfrac{dv}{dx} = 7x^6\).
\(\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{7^x\log 7}{7x^6}\) \(\left(= 7^{x-1}\dfrac{\log 7}{x^6}\right)\), for \(x \ne 0\).
