Implicit Differentiation
Implicit Differentiation – Introduction
Implicit differentiation is a technique used in calculus to differentiate equations that are not explicitly expressed in terms of one variable. It allows us to find the derivative of an equation that involves both dependent and independent variables. This concept is typically taught at the A Level Maths level.
Introduction
Implicit differentiation was first introduced by Isaac Newton and Gottfried Leibniz, the pioneers of calculus. They discovered that not all equations can be easily differentiated using the traditional methods. To overcome this limitation, they developed implicit differentiation as a powerful tool to differentiate more complex equations.
Understanding Implicit Differentiation
Implicit differentiation involves differentiating both sides of an equation with respect to the independent variable, treating the dependent variable as a function of the independent variable. This allows us to find the derivative of the equation without explicitly solving for the dependent variable.
Basic Examples
Let’s consider a few basic examples to understand how implicit differentiation works:
Example 1: Differentiating a Linear Equation
Consider the equation of a line, y = mx + c, where m and c are constants. To find the derivative of this equation, we differentiate both sides with respect to x:
dy/dx = d(mx + c)/dx
On the left-hand side, dy/dx represents the derivative of y with respect to x. On the right-hand side, we apply the derivative of a sum rule to differentiate mx and c:
dy/dx = m + 0
Therefore, the derivative of y with respect to x is simply equal to the constant m, which represents the slope of the line.
Example 2: Differentiating a Quadratic Equation
Let’s take the equation of a quadratic curve, x^2 + y^2 = r^2, where r is a constant. To differentiate this equation implicitly, we differentiate both sides with respect to x:
d(x^2 + y^2)/dx = d(r^2)/dx
On the left-hand side, we apply the derivative of a sum rule to differentiate x^2 and y^2 separately:
2x + 2y(dy/dx) = 0
Now, we can isolate dy/dx to find the derivative of y with respect to x:
This example demonstrates how implicit differentiation allows us to find the derivative of y with respect to x, even when y is not explicitly expressed as a function of x.
Implicit Differentiation - 2 Examples
\begin{aligned} u & =-3 x \quad v=y \\ \frac{d u}{d x} & =-3 \quad \frac{d v}{d x}=\frac{d y}{d x} \end{aligned}
\begin{aligned} 4 x-3 y-3 x \frac{d y}{d x}+2 y \frac{d y}{d x} & =0 \\ \frac{d y}{d x}(2 y-3 x) & =+3 y-4 x \\ \frac{d y}{d x} & =\frac{3 y-4 x}{2 y-3 x} \end{aligned}
At (1, -2)
\begin{aligned} \frac{d y}{d x} & =\frac{3(-2)-4(1)}{2(-2)-3(1)} \\ & =\frac{10}{7} \end{aligned}Gradient of normal is -\frac{7}{10}
\begin{aligned} y & =-\frac{7}{10} x+c \\ -2 & =-\frac{7}{10}(1)+c \\ c & =-\frac{13}{10} \\ y & =-\frac{7}{10} x-\frac{13}{10} \end{aligned}
\begin{array}{rlrl} u & =x & v=y \\ \frac{d x}{d x} & =1 & \frac{d v}{d x}=\frac{d y}{d x} \end{array}
\begin{aligned} 6 x+y+x \frac{d y}{d x}+2 y \frac{d y}{d x} & =0 \\ \frac{d y}{d x}(x+2 y) & =-6 x-y \\ \frac{d y}{d x} & =\frac{-6 x-y}{x+2 y} \end{aligned}
\begin{aligned} \frac{-6 x-y}{x+2 y} & =\frac{4}{3} \\ 3(-6 x-y) & =4(x+2 y) \\ -18 x-3 y & =4 x+8 y \\ 0 & =22 x+11 y \\ 0 & =2 x+y \end{aligned}
y=-2 x
\begin{aligned} 3 x^2+x(-2 x)+(-2 x)^2 & =20 \\ 3 x^2-2 x^2+4 x^2 & =20 \\ 5 x^2 & =20 \\ x^2 & =4 \\ x & = \pm 2 \end{aligned}
\begin{aligned} y & =-2(2) & y & =-2(-2) \\ & =-4 & & =4 \end{aligned}
(2,-4) \text { and }(-2,4)
Implicit Differentiation - 2 More Examples
\begin{array}{cc} u=4 x & v=y \\ \frac{d u}{d x}=4 & \frac{d v}{d x}=\frac{d y}{d x} \end{array}
\begin{aligned} 2 x+4 y+4 x \frac{d y}{d x}-1+2 y \frac{d y}{d x} & =0 \\ 4 x \frac{d y}{d x}+2 y \frac{d y}{d x} & =1-2 x-4 y \\ \frac{d y}{d x}(4 x+2 y) & =1-2 x-4 y \\ \frac{d y}{d x} & =\frac{1-2 x-4 y}{4 x+2 y} \end{aligned}
At (2, 3)
\begin{aligned} \frac{d y}{d x} & =\frac{1-2(2)-4(3)}{4(2)+2(3)} \\ & =\frac{-15}{14} \end{aligned}\begin{aligned} & y=\frac{-15}{14} x+c \\ & 3=-\frac{15}{14}(2)+c \\ & c=\frac{36}{7} \\ & y=-\frac{15}{14} x+\frac{36}{7} \end{aligned}
\begin{aligned} 2 \cos x & =2 \sin y \frac{d y}{d x} \\ \frac{2 \cos x}{2 \sin y} & =\frac{d y}{d x} \\ \frac{d y}{d x} & =\frac{\cos x}{\sin y} \end{aligned}
\begin{aligned} \frac{\cos x}{\sin y} & =0 \\ \cos x & =0 \\ x & =\frac{1}{2} \pi \end{aligned}
\begin{aligned} 2 \sin \left(\frac{1}{2} \pi\right)+2 \cos y & =3 \\ 2+2 \cos y & =3 \\ 2 \cos y & =1 \\ \cos y & =1 / 2 \\ y & =\frac{\pi}{3} \end{aligned}
\left(\frac{\pi}{2}, \frac{\pi}{3}\right)
Implicit Differentiation – Final Example
\begin{aligned} y & =2^x \\ \ln y & =\ln 2^x \\ \ln y & =x \ln 2 \\ \frac{1}{y} \frac{d y}{d x} & =\ln 2 \\ \frac{d y}{d x} & =y \ln 2 \\ & =2^x \ln 2 \end{aligned}
y=3^{x^2}
\begin{aligned} \ln y & =\ln 3^{x^2} \\ \ln y & =x^2 \ln 3 \\ \frac{1}{y} \frac{d y}{d x} & =2 x \ln 3 \\ \frac{d y}{d x} & =y 2 x \ln 3 \\ & =3^{x^2} \cdot 2 x \ln 3 \end{aligned}
Conclusion
Implicit differentiation is a powerful technique used in calculus to differentiate equations that are not explicitly solved for one variable. It was introduced by Newton and Leibniz to overcome the limitations of traditional differentiation methods. By differentiating both sides of an equation with respect to the independent variable, treating the dependent variable as a function, we can find the derivative of the equation. This concept is fundamental in higher-level mathematics and serves as a foundation for more complex applications of calculus.