Question

If \( x = f(t) \) and \( y = g(t) \) are differentiable functions of \( t \) so that \( y \) is a function of \( x \) and if \( \frac{dx}{dt} \neq 0 \), then prove that \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \] Hence, find the derivative of \( 7^x \) with respect to \( x^7 \).

Hint:

To find the derivative of a function with respect to a different variable, use the chain rule and account for the derivative of the intermediate variable.

Answer 1

Step 1: Chain rule proof.
By the chain rule, if \( y \) is a function of \( t \) and \( x \) is also a function of \( t \), then: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] This follows from the fact that \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) are the rates of change of \( y \) and \( x \) with respect to \( t \), and their ratio gives the rate of change of \( y \) with respect to \( x \).

Step 2: Find the derivative of \( 7^x \) with respect to \( x^7 \).
We want to find \( \frac{d}{dx^7} \) of \( 7^x \). First, apply the chain rule: \[ \frac{d}{dx^7}(7^x) = \frac{d}{dx}\left( 7^x \right) \times \frac{d}{dx}\left( x^7 \right) \] The derivative of \( 7^x \) with respect to \( x \) is \( 7^x \ln 7 \), and the derivative of \( x^7 \) with respect to \( x \) is \( 7x^6 \). Thus, the result is: \[ \frac{d}{dx^7}(7^x) = 7^x \ln 7 \times 7x^6 = 7^{x+1} \ln 7 \times x^6 \]

Final Answer: The derivative of \( 7^x \) with respect to \( x^7 \) is: \[ \boxed{7^{x+1} \ln 7 \times x^6} \]