Question

In Exercises \(15-22,\) find \(d y / d x\) . Support your answer graphically. $$y=\frac{x^{2}}{1-x^{3}}$$

Short answer

The derivative of the given function is \(\frac{d y}{d x} =\frac{2x - 5x^{5}}{(1-x^{3})^{2}}\).

Step-by-step solution

Understanding the Quotient Rule

The quotient rule states that the derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2}\). Now, applying the quotient rule to the given function using \(f(x)=x^{2}\) and \(g(x)=1-x^{3}\).

Differentiating

The derivative of \(f(x)=x^{2}\) is \(f'(x)= 2x\), while the derivative of \(g(x)=1-x^{3}\) is \(g'(x)= -3x^{2}\). Computing the numerator, we multiply f'(x) with g(x) and subtract from it the product of g'(x) and f(x), thus we get \(f'(x)g(x) - g'(x)f(x)= 2x(1-x^{3}) - (-3x^{2})(x^{2}) = 2x - 5x^{5}\).

Implementing Quotient Rule

We implement the quotient rule which yields \(\frac{d y}{d x} =\frac{2x - 5x^{5}}{(1-x^{3})^{2}}\).

Graphical Representation

Graph the given function y and its derivative dy/dx on the same axes. The curve of the function y generally moves upwards as x values increase. The curve representing dy/dx, on the other hand, shows the slope of the tangent line to the curve at any given point, meaning it represents the rate at which y is changing at each x value. A higher value dy/dx corresponds to a steeper slope of the y function at that point, and the opposite for a lower dy/dx value. When dy/dx is 0, the y function has a horizontal tangent line.