Implicit Differentiation: Videos & Practice Problems

Implicit differentiation is a technique used to find the derivative of a variable, y, that is not isolated on one side of an equation. To differentiate, take the derivative of each term with respect to x, applying the chain rule for terms involving y. The result is expressed as $dy/dx$ = -x/y. This method is particularly useful when y is complex or appears multiple times, allowing for efficient differentiation without needing to solve for y first.

Finding The Implicit Derivative

In calculus, understanding how to differentiate equations where the variable \( y \) is embedded within the equation, rather than isolated, is crucial. This technique is known as implicit differentiation. The process begins by taking the derivative of each term in the equation with respect to \( x \). For instance, if you have an equation like \( x^2 + y^2 = 49 \), you would differentiate both sides with respect to \( x \).

Using the power rule, the derivative of \( x^2 \) is \( 2x \), and since \( 49 \) is a constant, its derivative is \( 0 \). The challenge arises when differentiating \( y^2 \). Here, the chain rule comes into play. The chain rule 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 \( y^2 \) with respect to \( x \) is \( 2y \cdot \frac{dy}{dx} \).

After differentiating, the equation becomes \( 2x + 2y \cdot \frac{dy}{dx} = 0 \). To isolate \( \frac{dy}{dx} \), you would first rearrange the equation to get \( 2y \cdot \frac{dy}{dx} = -2x \). Dividing both sides by \( 2y \) gives \( \frac{dy}{dx} = -\frac{x}{y} \). 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, you can solve the original equation for \( y \). For the example above, rearranging gives \( y = \sqrt{49 - x^2} \). Substituting this back into the derivative yields \( \frac{dy}{dx} = -\frac{x}{\sqrt{49 - x^2}} \). This demonstrates that both methods—implicit differentiation and explicit differentiation after solving for \( y \)—yield the same result.

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 enhances your ability to tackle a wider range of calculus problems efficiently.

Find $dy/dx$ for the equation below using implicit differentiation.
$5x^2+y^3=12$

• A. $dy/dx=-\frac{10x}{3y^2}$
• B. $dy/dx=\frac{10x}{3y^2}$
• C. $dy/dx=\left(12-5x^2\right)^{\left(\frac13\right)}$
• D. $dy/dx=-\left(12-5x^2\right)^{\left(\frac13\right)}$

Answer: (A)

Find $dy/dx$ for the equation below using implicit differentiation.
$3\sqrt{y}=x-y$

• A. $dy/dx=3\sqrt{y}-x$
• B. $dy/dx=\frac{2\sqrt{y}-3}{4\sqrt{y}}$
• C. $dy/dx=\frac{2\sqrt{y}}{3+2\sqrt{y}}$
• D. $dy/dx=\frac{3}{3\sqrt{y}-2xy}$

Answer: (C)

Find $dy/dx$ for the equation below using implicit differentiation.
$xy=\sin\left(y\right)$

• A. $dy/dx=\frac{\sin\left(y\right)}{x}$
• B. $dy/dx=\frac{x}{\cos\left(y\right)}$
• C. $dy/dx=\frac{y}{\cos\left(y\right)-x}$
• D. $dy/dx=\cos\left(y\right)$

Answer: (C)

Finding The Implicit Derivative Example 1

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 us:

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

Multiplying both sides by \( y^2 \) leads to:

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

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 moving \( \frac{2}{x} \) to the right side:

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

Multiplying through by \( -1 \) gives:

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

Taking the reciprocal results in:

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

To simplify further, 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 this 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, we have successfully demonstrated the process of implicit differentiation and explicit differentiation, along with verifying the consistency of our findings.

Here’s what students ask on this topic:

What is implicit differentiation and when is it used?

Implicit differentiation is a technique used to find the derivative of a variable, y, that is not isolated on one side of an equation. Instead of solving for y explicitly, you differentiate each term with respect to x, applying the chain rule for terms involving y. This method is particularly useful when y is complex or appears multiple times in the equation, allowing for efficient differentiation without needing to solve for y first. For example, in the equation x² + y² = 49, implicit differentiation helps find dy/dx without isolating y.

How do you apply the chain rule in implicit differentiation?

To apply the chain rule in implicit differentiation, identify the outer and inner functions. For a term like y², the outer function is the square, and the inner function is y. Differentiate the outer function with respect to the inner function, yielding 2y. Then, multiply by the derivative of the inner function with respect to x, which is dy/dx. Thus, the derivative of y² with respect to x is 2y(dy/dx). This process is repeated for all terms involving y in the equation.

Can you provide an example of implicit differentiation?

Sure! Consider the equation x² + y² = 49. To find dy/dx, differentiate both sides with respect to x: d/dx(x²) + d/dx(y²) = d/dx(49). Using the power rule, the derivative of x² is 2x. For y², apply the chain rule: 2y(dy/dx). The derivative of the constant 49 is 0. So, we have 2x + 2y(dy/dx) = 0. Solving for dy/dx, we get dy/dx = -x/y.

Why is implicit differentiation useful in calculus?

Implicit differentiation is useful in calculus because it allows for the differentiation of equations where y is not easily isolated. This technique is particularly helpful when dealing with complex equations or when y appears multiple times. It simplifies the process by avoiding the need to solve for y explicitly, making it more efficient and practical for finding derivatives in such scenarios.

What are the steps to perform implicit differentiation?

The steps to perform implicit differentiation are: 1) Differentiate each term of the equation with respect to x. 2) Apply the chain rule to terms involving y, multiplying by dy/dx. 3) Collect all dy/dx terms on one side of the equation. 4) Solve for dy/dx. This process allows you to find the derivative of y with respect to x without isolating y first.