Question

1. If y = -Ax^3/5, Find dy/dx, where A is a constant
2. If y = 1/x, find dy/dx using first principle.

Answer 1

1. If $y = -Ax^{\frac{3}{5}}$. Find $\frac{dy}{dx}$, where $A$ is a constant

2. If $y = \frac{1}{x}$, find $\frac{dy}{dx}$ using first principle.

**Solution:**

Using that derivatives of powers: if $f(x) = x^n$, where $n$ is any real number, then $f'(x) = nx^{n-1}$

So

If $y = -Ax^{\frac{3}{5}}$, and $A$ is a constant

$$y' = \left(-A x^{\frac{3}{5}}\right)' = -A \left(x^{\frac{3}{5}}\right)' = -A * \frac{3}{5} * x^{\frac{3}{5} - 1} = -\frac{3A}{5} x^{-\frac{2}{3}}$$

**Answer 1:** $y' = -\frac{3A}{5} x^{-\frac{2}{3}}$

From the definition of the derivative $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{y(x + \Delta x) - y(x)}{\Delta x}$

If $y = \frac{1}{x}$ we have:

$$\begin{array}{l} y'(x) = \lim_{\Delta x \to 0} \frac{y(x + \Delta x) - y(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{1}{x + \Delta x} - \frac{1}{x}}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{x - x - \Delta x}{(x + \Delta x) * x}}{\Delta x} = \lim_{\Delta x \to 0} \frac{-\Delta x}{\Delta x * (x + \Delta x) * x} \\ = \lim_{\Delta x \to 0} \frac{-1}{(x + \Delta x) * x} = -\frac{1}{x^2} \end{array}$$

**Answer 2:** $y' = -\frac{1}{x^2}$