Find the derivative of $y = \frac{x^2 - 16}{x^3}$ using the four step process (Definition of Derivatives)

Question

I want to see how it is solved using the four-step process

Oh what I mean by the four-step process is the increment method;

1. Replace by +Δ and by +Δ.
2. Solve for Δ
3. By some suitable transformation, change the right member of the equation in Step 2 into a form which contains Δ explicitly as a factor and divide through by Δ.
4. Determine /=Δ→ [Δ/Δ]

When I try to solve it, I don't know how to find a way to eliminate Δ in the denominator, making it undefined.

Answer 1

We have $$y=\frac{1}{x}-\frac{16}{x^3}$$ Let us introduce an increment of $\Delta x$ in $x$ which results in a change $\Delta y$ in $y$. This implies $$ y+\Delta y= \frac{1}{x+\Delta x}-\frac{16}{(x+\Delta x)^3} $$ Subtracting the two equations we get $$ \Delta y = \frac{1}{x+\Delta x}- \frac{1}{x} +\frac{16}{x^3}-\frac{16}{(x+\Delta x)^3}$$ $$\implies \Delta y = \frac{-\Delta x }{x(x+\Delta x)} + \frac{16(\Delta x^3+3x \Delta x^2+ 3x^2\Delta x)}{x^3(x+\Delta x)^3}$$ Therefore, we get $$\frac{\Delta y }{\Delta x } = \frac{-1 }{x(x+\Delta x)} + \frac{16(\Delta x^2+3x \Delta x+ 3x^2)}{x^3(x+\Delta x)^3}$$ Taking the limit $\Delta x\to 0$ we get, $$ \frac{dy}{dx} =\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0} \frac{-1 }{x(x+\Delta x)} + \frac{16(\Delta x^2+3x \Delta x+ 3x^2)}{x^3(x+\Delta x)^3} = -\frac{1}{x^2}+\frac{16\cdot 3x^2}{x^6} = -\frac{1}{x^2}+\frac{48}{x^4}$$