![]() ![]() Inside the cosine, we have the product, x √(x 2 – 10), to which we will be applying the product rule to find its derivative (notice that we are always moving from the outside to the inside, in order to discover the operation that needs to be tackled next):ĭu / dx = ( x √(x 2 – 10) )’ = √(x 2 – 10) + x ( √(x 2 – 10) )’ In finding its derivative by the chain rule, we shall be using the intermediate variable, u:ĭh / du = (cos( x √( x 2 – 10) ))’ = -sin( x √( x 2 – 10) ) We find that the outermost function is a cosine. If we observe this closely, we realize that not only do we have nested functions for which we will need to apply the chain rule multiple times, but we also have a product to which we will need to apply the product rule. Putting the two parts together, we obtain the derivative of the composite function:ĮXAMPLE 3: Let’s now raise the bar a little further by considering a more challenging composite function: Finding its derivative, again ignoring the inside, gives us:ĭh / du = (cos( x 3 – 1))’ = -sin( x 3 – 1) The outer function in this case is, cos x. We will again use, u, the output of the inner function, as our intermediate variable. Putting the two parts together and simplifying, we have:ĮXAMPLE 2: Let’s repeat the procedure, this time with a different composite function: The next step is to find the derivative of the inner part of the composite function, this time ignoring whatever is outside. For this purpose, we can apply the power rule: The first step is to find the derivative of the outer part of the composite function, while ignoring whatever is inside. The output of the inner function is denoted by the intermediate variable, u, and its value will be fed into the input of the outer function. We can separate the composite function into the inner function, f( x) = x 2 – 10, and the outer function, g( x) = √ x = ( x) 1/2. We will be starting with univariate functions first, and then apply what we learn to multivariate functions.ĮXAMPLE 1: Let’s raise the bar a little by considering the following composite function: Let’s see a few more challenging ones here. We have already discovered the chain rule for univariate and multivariate functions, but we have only seen a few simple examples so far. The Chain Rule on Multivariate Functions.This tutorial is divided into two parts they are: Applying the chain rule to multivariate functions requires the use of partial derivatives. ![]() Along the way, the application of other derivative rules might be required. The application of the chain rule follows a similar process, no matter how complex the function is: take the derivative of the outer function first, and then move inwards.The process of applying the chain rule to univariate functions can be extended to multivariate ones.In this tutorial, you will discover how to apply the chain rule of calculus to challenging functions.Īfter completing this tutorial, you will know: ![]() We will be building on our earlier introduction to the chain rule, by tackling more challenging functions. It is essential in understanding the workings of the backpropagation algorithm, which applies the chain rule extensively in order to calculate the error gradient of the loss function with respect to each weight of a neural network. The chain rule is an important derivative rule that allows us to work with composite functions. ![]()
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