Implicit differentiation

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.

\[\frac{d}{dx}(uv) = v\frac{d}{dx}(u) + u \frac{d}{dx}(v)\]

\[\frac{d}{dx}(xy) = y\frac{d}{dx}(x) + x \frac{d}{dx}(y)\]

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 \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = \frac{y-5x^4}{2y-x}\]

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:

\[\frac{d^2y}{dx^2} = -1\]

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.

\(y-y_1 = a(x-x_1) \Rightarrow y-1 = 6(x-1) \Rightarrow y = -6x + 7\)

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.

\(y - y_1 = \frac{dy}{dx}(x - x_1)\)

\(y -1 = \frac{1}{6}(x -1) \Rightarrow y = \frac{x}{6} + \frac{5}{6}\)