Question

If x = f(t) and y = g(t) are differentiable functions of t so that y is a function of x and if `(dx)/(dt)` ≠ 0 then prove that `(dy)/(dx) = ((dy)/(dt))/((dx)/(d"))`.

Hence, find the derivative of 7^x w.r.t. x⁷.

Answer 1

`dy/dx = ((dy/dt)/(dx/dt))`

We are given that x = f(t) and y = g(t) are differentiable functions of t, meaning we can express y as a function of x through t.

By the Chain Rule, the derivative of y with respect to x can be written as:

`dy/dx = dy/dt xx dt/dx`

Since `(dx)/(dt) ≠ 0`, we can take its reciprocal:

`(dt)/(dx) = 1/((dx)/(dt))`

`(dy)/(dx) = (dy)/(dt) xx 1/((dx)/(dt)) = (dy/dt)/(dx/dt)`

`(dy)/(dx) = (dy/dt)/(dx/dt)`

∴ The derivative of 7^x w.r.t. x⁷

`d/(du)(7^x)`, where u = x⁷

Using the Chain Rule:

`d/(du)(7^x) = (d/dx(7^x))/(d/dx(x^7))`

Using the exponential differentiation formula:

∴ `(dy)/(du) = (7^x ln 7)/(7x^6)`

∴ `(dy)/(du) = (7^(x - 1) ln 7)/(x^6)`