Implicit Derivatives

Implicit Differentiation is the process of differentiating an expression where \(y\) isn't isolated. It's commonly used when it's difficult or impossible to solve for \(y\).

You assume \(y\) can be solved in terms of \(x\) and treat it as a function in terms of \(x\). Therefore, you must apply Chain Rule since you're assuming \(y\) is defined in terms of \(x\).

The Chain Rule states:

\(\cfrac{dy}{dx} = \cfrac{dy}{du}\cdot \cfrac{du}{dx}\)

This shows that \(y\) is expressed in terms of \(u\) and that \(u\) is expressed in terms of \(x\). These expressions can be written as \(y = y(u)\) and \(u = u(x)\) respectively.

It's important to take different variables into account when working with Chain Rule and Implicit Differentiation by extension. For example, if we wanted to express \(p\) in terms of \(y\), we would write it as:

\(p = p(y)\)

Likewise, if wanted to express \(y\) in terms of \(x\), we would write it as:

\(y = y(x)\)

Therefore, we could express this in Chain Rule as:

\(\cfrac{dp}{dx} = \cfrac{dp}{dy}\cdot \cfrac{dy}{dx}\)

NOTE: \(\cfrac{dy}{dx}\) can also be expressed as \(f'\).


Example

Differentiate \(p = 2y^5\) using Implicit Differentiation.

First, we can determine \(\cfrac{dp}{dy}\):

\(\cfrac{dp}{dy} = 10y^4\)

As you can see, we have differentiated \(2y^5\) with respect to \(y\).

Next, we can determine the full derivative using Chain Rule:

\(\cfrac{dp}{dx} = \cfrac{dp}{dy}\cdot \cfrac{dy}{dx}\)

\(\cfrac{dp}{dx} = 10y^4\cdot \cfrac{dy}{dx}\)

Therefore, we can determine that \(p = 2y^5\) differentiated using Implicit Differentiation is \(\cfrac{dp}{dx} = 10y^4\cdot \cfrac{dy}{dx}\).


Differentiate the equation \(2x^2 + y^2 =4\) using both Implicit and Explicit Differentiation.

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