Implicit Differentiation: Videos & Practice Problems

In calculus, understanding how to differentiate equations where the variable \( y \) is not isolated on one side is crucial. This situation often arises in implicit differentiation, a technique used to find the derivative of \( y \) with respect to \( x \) when \( y \) is embedded within an equation. The process begins by taking the derivative of each term in the equation with respect to \( x \).

For example, consider the equation \( x^2 + y^2 = 49 \). To differentiate, apply the derivative operator to both sides. The derivative of \( x^2 \) is \( 2x \), and since \( 49 \) is a constant, its derivative is \( 0 \). The challenge lies in differentiating \( y^2 \). Here, the chain rule comes into play, which states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. Thus, the derivative of \( y^2 \) becomes \( 2y \cdot \frac{dy}{dx} \).

After differentiating, the equation becomes:

\( 2x + 2y \cdot \frac{dy}{dx} = 0 \).

To isolate \( \frac{dy}{dx} \), rearrange the equation:

\( 2y \cdot \frac{dy}{dx} = -2x \).

Dividing both sides by \( 2y \) yields:

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

This result represents the implicit derivative. In some cases, you may need to express \( \frac{dy}{dx} \) solely in terms of \( x \). To do this, solve the original equation for \( y \). From \( x^2 + y^2 = 49 \), rearranging gives:

\( y = \sqrt{49 - x^2} \).

Substituting this expression for \( y \) back into the derivative results in:

\( \frac{dy}{dx} = -\frac{x}{\sqrt{49 - x^2}} \).

This method of implicit differentiation is particularly useful when dealing with complex equations where isolating \( y \) is not straightforward, such as when \( y \) appears in higher powers or multiple terms. Mastering this technique allows for efficient differentiation in a variety of mathematical contexts.

To find the derivative of the equation \( \frac{2}{x} - \frac{1}{y} = 36 \) using implicit differentiation, we start by differentiating each term with respect to \( x \). The first term, \( \frac{2}{x} \), can be rewritten as \( 2x^{-1} \). Applying the power rule, the derivative becomes \( -\frac{2}{x^2} \). For the second term, \( \frac{1}{y} \), we rewrite it as \( y^{-1} \). The derivative of this term, using the chain rule, is \( -\frac{1}{y^2} \cdot \frac{dy}{dx} \) or \( -\frac{1}{y^2} y' \). The derivative of the constant \( 36 \) is \( 0 \). Thus, we have:

\[-\frac{2}{x^2} - \frac{1}{y^2} y' = 0\]

Rearranging gives:

\[\frac{1}{y^2} y' = \frac{2}{x^2}\]

Multiplying both sides by \( y^2 \) results in:

\[y' = \frac{2y^2}{x^2}\]

This expression represents the implicit derivative \( y' \) in terms of both \( x \) and \( y \).

Next, we solve the original equation explicitly for \( y \). Starting from:

\[\frac{2}{x} - \frac{1}{y} = 36\]

We isolate \( \frac{1}{y} \) by subtracting \( \frac{2}{x} \) from both sides:

\[-\frac{1}{y} = 36 - \frac{2}{x}\]

Multiplying through by \( -1 \) gives:

\[\frac{1}{y} = \frac{2}{x} - 36\]

Taking the reciprocal yields:

\[y = \frac{1}{\frac{2}{x} - 36}\]

To simplify, we can multiply the numerator and denominator by \( x \) to obtain a common denominator:

\[y = \frac{x}{2 - 36x}\]

Now, we differentiate this expression for \( y \) using the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then:

\[y' = \frac{u'v - uv'}{v^2}\]

Here, \( u = x \) and \( v = 2 - 36x \). The derivatives are \( u' = 1 \) and \( v' = -36 \). Applying the quotient rule gives:

\[y' = \frac{(1)(2 - 36x) - (x)(-36)}{(2 - 36x)^2}\]

Expanding the numerator results in:

\[y' = \frac{2 - 36x + 36x}{(2 - 36x)^2} = \frac{2}{(2 - 36x)^2}\]

This expression for \( y' \) is now in terms of \( x \) only.

Finally, we check the consistency of our results by substituting the expression for \( y \) back into the implicit derivative we found earlier. Substituting \( y = \frac{x}{2 - 36x} \) into \( y' = \frac{2y^2}{x^2} \) gives:

\[y' = \frac{2\left(\frac{x}{2 - 36x}\right)^2}{x^2} = \frac{2 \cdot \frac{x^2}{(2 - 36x)^2}}{x^2} = \frac{2}{(2 - 36x)^2}\]

This matches the result obtained from differentiating \( y \) explicitly, confirming that both methods yield consistent results. Thus, the implicit derivative and the explicit derivative are verified to be correct.