Applications of differentiation

Applications of differentiation

Differentiation can be applied in many areas. Some common applications include finding:

1. Slope of curves
2. Rate of change
3. Critical points of a curve
4. Marginal cost
5. Marginal revenue

Slope of a curve at a given point

The slope of a curve at a given point is defined as the slope of the tangent line to the curve at that point. It measures the rate of increase (or decrease) of $y$ with respect to $x$. The slope of the curve is always equal to the slope of the tangent at the point of contact.

Consider a curve $y = f(x)$ and a point $P(x_1, y_1)$ on the curve.

Figure 4.2: Illustration showing the slope of a curve at a given point

The slope $m$ at $x = x_1$ is given by the derivative:

$$m = \left. \frac{dy}{dx} \right|_{x = x_1} = f'(x_1)$$

Example

Find the slope of the curve $y = x^2$ at the point $(4, 16)$.

Solution:

The slope is given by:

$$m = \frac{dy}{dx}$$

Differentiating $y = x^2$ with respect to $x$:

$$\frac{dy}{dx} = 2x$$

At the point $(4, 16)$, substitute $x = 4$:

$$m = 2 \times 4 = 8$$

Therefore, the slope of the curve at $(4, 16)$ is 8.

Example

Calculate the gradient of the tangent to the curves:

1. $y = x^2 - 2x + 1$ at $x = 2$
2. $y = (x+2)(x-4)$ at $x = 3$

Solution:

a. Given $y = x^2 - 2x + 1$, differentiate:

$$\frac{dy}{dx} = 2x - 2$$

At $x = 2$:

$$m = 2(2) - 2 = 4 - 2 = 2$$

Thus, the gradient at $x = 2$ is 2.

b. Given $y = (x+2)(x-4)$, first expand:

$$y = x^2 - 2x - 8$$

Differentiating:

$$\frac{dy}{dx} = 2x - 2$$

At $x = 3$:

$$m = 2(3) - 2 = 6 - 2 = 4$$

Therefore, the gradient at $x = 3$ is 4.