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Implicit differentiation can also be used to describe the slope and concavity of curves which are defined by the parametric equations. These types of equations often describe curves of implicit functions. The derivatives can then be used to construct the equations of tangents and normal lines of curves.
How do we carry out implicit differentiation?
For explicit functions of the form \(y = ax+\)... we perform standard differentiation using the chain rule. However, if the functions are written implicitly in the form \(x + y = a\), then we use a variation of the chain rule due to the assumption that y can be expressed as a function of x.
We differentiate all the variables for both sides of the equation (left side and right side) with respect to x. However, for y terms, we will multiply by \(\frac{dy}{dx}\).
After we differentiate, we solve for \(\frac{dy}{dx}\).
A circle is described by the equation below. Find the derivative of the circle.
\[x^2 + y^2 = 49\]
Solution:
We first differentiate each part of the equation. However, for the y term, we also need to multiply by dy/dx.
\[\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(49)\]
\[2x + 2y \frac{dy}{dx} = 0\]
\[\frac{dy}{dx} = - \frac{2x}{2y}\]
Find \(\frac{dy}{dx}\) if
\[x^5 + y^2 - x \cdot y = 9\]
Solution:
\[\frac{d}{dx}(x^5) + \frac{d}{dx}(y^2) - \frac{d}{dx}(xy) = \frac{d}{dx}(9) = 0\]
In this example, we also have the multiplication of two variables which means we have to also include the Product Rule for differentiating the term xy. The formula for the product rule is the one below.
\[\frac{d}{dx}(uv) = v\frac{d}{dx}(u) + u \frac{d}{dx}(v)\]
Hence the formula for the derivative of the xy term is:\[\frac{d}{dx}(xy) = y\frac{d}{dx}(x) + x \frac{d}{dx}(y)\]
If we follow the procedure below we get:
\[5x^4 + 2y \frac{dy}{dx} - (1y + x\frac{dy}{dx}) = 0\]
\[5x^4 + 2y\frac{dy}{dx} - y -x\frac{dy}{dx} = 0\]
\[\frac{dy}{dx}(2y - x) = y - 5x^4\]
And solving for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{y-5x^4}{2y-x}\]
Higher order implicit differentiation.
For higher order differentiation we proceed with the same process; however, in order to find the second derivative we need to differentiate the first derivative, and in order to find the third derivative we must differentiate the second derivative and so on. We can generalise this using the formula.
Here n is the order of the derivative.
\[\frac{d^ny}{dx^n} = \frac{d}{dx} (\frac{d^{n-1}y}{dx^{n-1}})\]
In order to differentiate implicit functions, we continue with the implicit differentiation rules mentioned earlier in order to find the lower order derivative first. To find the higher order derivatives we need to also apply the above formula.
Find the second derivative of the following expression.
\[x^2 + 2y = 3\]
Differentiating with respect to x:
\[2x + 2\frac{dy}{dx} = 0\]
\[\frac{dy}{dx} = -x\]
Differentiating again with respect to x:
\[\frac{d^2y}{dx^2} = -1\]
How do we find the equation of the tangent of a curve using implicit differentiation?
We can find the equation of a tangent – which is a straight line – using the implicit differentiation procedure to find the slope of a curve. If the point of tangency is known, the equation of a straight line can be found using the formula below, where x1,y1 are the coordinates of the point and a is the slope or derivative of the curve.
\[y - y_1 = a(x-x_1)\]
A curve has an equation as below.
\[x^3 + x^2 + xy = 3\]
Find the equation of the tangent at the point where x = 1.
We start by finding the point of tangency. We can do this by substituting the coordinate given, to find the possible y coordinate.
\[1^3 + 1^2 + 1y = 3 \Rightarrow y = 1\]
To determine the equation of the tangent we also need the slope of the curve at the point of tangency.
\[\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2)+ \frac{d}{dx}(xy) = \frac{d}{dx} (3)\]
\(3x^2 + 2x + y+ x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-3x^2 - 2x -y}{x} = \frac{-3 \cdot 1^2 - 2 \cdot 1 -1}{1} = 6\)
We can construct the equation of the tangent using the formula given.
\(y-y_1 = a(x-x_1) \Rightarrow y-1 = 6(x-1) \Rightarrow y = -6x + 7\)
How do we find the equation of the normal to a curve using implicit differentiation?
Similarly, we follow the same process to find the equation of the normal to a curve. However, the slope of the normal is always the negative reciprocal of the slope of the tangent of the curve as shown in the formula below where N and T indicate the slope of the normal and tangent respectively.
\(slope_N \cdot Slope_T = -1\)
Find the slope of the normal to the curve given in the previous example.
Solution:
Continuing the previous example, the slope of the normal can be found by using the derivative that was found.
\(\frac{dy}{dx} = -6\)
Using the formula for the slope of the normal to the curve, the slope of the normal is the negative reciprocal of the slope of the curve since it is perpendicular to the curve. Hence the slope of the normal is 1/6. Using the formula for the construction of a straight line, the equation of the normal is as follows.
\(y - y_1 = \frac{dy}{dx}(x - x_1)\)
\(y -1 = \frac{1}{6}(x -1) \Rightarrow y = \frac{x}{6} + \frac{5}{6}\)
Implicit Differentiation - Key takeaways
Implicit differentiation is a method that is used when both unknown variables are used in an equation not isolated on one side of the equation.
All terms are differentiated and the y term needs to be multiplied by \(\frac{dy}{dx}\).
Higher order implicit differentiation is used when a second or third derivative is needed. The lower order derivative is differentiated to find the order of derivative required.
The equation of a tangent and normal to a curve described by implicit equations can be found using the implicit derivative and the coordinates.
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Frequently Asked Questions about Implicit differentiation
How do you differentiate implicit functions??
The terms of the equation are differentiated in terms of x as normal, but the y term needs to also be multiplied by dy/dx.
What is implicit differentiation ?
Implicit differentiation is a method that allows differentiation of y with respect to x (dy/dx) without the need of solving for y.
When do you use implicit differentiation?
Implicit differentiation is used when implicit functions are present where both unknown variables are part of an equation, and none are isolated on one side of the equation.
How to find dy/ dx by implicit differentiation?
The dy/dx term is found from the differentiation from the y term which is multiplied by dy/dx. Then this term needs to be isolated on one side to find the derivative expression.
Why is implicit differentiation used?
It allows differentiation of y with respect to x (dy/dx) without the need of solving for y.
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