When dy/dx > 0 as x increases…
When dy/dx > 0 as x increases, it means that the derivative of the function y with respect to x is positive for increasing values of x
When dy/dx > 0 as x increases, it means that the derivative of the function y with respect to x is positive for increasing values of x. In other words, the slope of the tangent line to the curve of the function y is positive for increasing x values.
To understand this concept, let’s look at an example. Consider the function y = x^2. Taking the derivative of y with respect to x, we get dy/dx = 2x.
When x is positive and increasing, the value of dy/dx = 2x is also positive. This indicates that as x increases, the slope of the tangent line to y = x^2 at any given point also increases. Visually, this means that the curve of the function y = x^2 gets steeper and steeper as x increases.
Similarly, if we consider a function like y = -x^3, taking its derivative gives us dy/dx = -3x^2. In this case, when x is positive and increasing, the value of dy/dx = -3x^2 is negative. This indicates that as x increases, the slope of the tangent line to y = -x^3 decreases, resulting in a curve that gets less steep as x increases.
Essentially, when dy/dx > 0 as x increases, it implies that the function is increasing, and the curve becomes steeper as x increases. Conversely, when dy/dx < 0 as x increases, the function is decreasing, and the curve becomes less steep as x increases.
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