The Chain Rule: Videos & Practice Problems

To find the derivative of a composite function, such as \( (4x + 5)^3 \), we can utilize the chain rule, which simplifies the process significantly compared to expanding the expression. The chain rule states that to differentiate a function that is composed of two functions, we start from the outside and work our way in.

For a function expressed as \( f(g(x)) \), the derivative is given by:

\[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]

In our example, \( f(u) = u^3 \) where \( u = 4x + 5 \). First, we differentiate the outer function:

\[f'(u) = 3u^2\]

Next, we substitute back the inner function:

\[f'(g(x)) = 3(4x + 5)^2\]

Now, we need to find the derivative of the inner function \( g(x) = 4x + 5 \):

\[g'(x) = 4\]

Combining these results using the chain rule gives us:

\[\frac{d}{dx} (4x + 5)^3 = 3(4x + 5)^2 \cdot 4 = 12(4x + 5)^2\]

In another example, consider the function \( f(x) = 2(3x^2 - x)^4 \). Here, we identify the outer function as \( f(u) = 2u^4 \) and the inner function as \( g(x) = 3x^2 - x \). We start by differentiating the outer function:

\[f'(u) = 8u^3\]

Substituting back gives us:

\[f'(g(x)) = 8(3x^2 - x)^3\]

Next, we differentiate the inner function:

\[g'(x) = 6x - 1\]

Applying the chain rule results in:

\[f'(x) = 8(3x^2 - x)^3 \cdot (6x - 1)\]

In summary, the chain rule is a powerful tool for finding derivatives of composite functions efficiently. By identifying the outer and inner functions, differentiating them separately, and then combining the results, we can simplify the differentiation process significantly. Regular practice with these concepts will enhance your understanding and proficiency in calculus.

To find the derivative \(\frac{dy}{dx}\) of the function \(y = (2x - 1)^4(3 + x^2)\), we need to apply both the product rule and the chain rule due to the presence of functions within functions and their multiplication.

First, we recognize that the product rule states that if we have two functions \(u\) and \(v\), the derivative is given by:

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

In our case, let \(u = (2x - 1)^4\) and \(v = 3 + x^2\). We will differentiate each function separately.

Starting with \(u\), we apply the chain rule. The chain rule states that if \(y = f(g(x))\), then:

\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]

Here, the outer function is \(f(g) = g^4\) and the inner function is \(g(x) = 2x - 1\). The derivative of the outer function is:

\[f'(g) = 4g^3\]

Substituting back, we have:

\[f'(2x - 1) = 4(2x - 1)^3\]

The derivative of the inner function \(g(x)\) is:

\[g'(x) = 2\]

Thus, the derivative of \(u\) is:

\[\frac{du}{dx} = 4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3\]

Next, we differentiate \(v = 3 + x^2\). The derivative is straightforward:

\[\frac{dv}{dx} = 0 + 2x = 2x\]

Now we can apply the product rule:

\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]

Substituting in our values:

\[\frac{dy}{dx} = (2x - 1)^4(2x) + (3 + x^2)(8(2x - 1)^3)\]

Now, we simplify this expression. The first term is:

\[2x(2x - 1)^4\]

For the second term, we distribute:

\[8(2x - 1)^3(3 + x^2) = 24(2x - 1)^3 + 8x^2(2x - 1)^3\]

Combining these, we have:

\[\frac{dy}{dx} = 2x(2x - 1)^4 + 24(2x - 1)^3 + 8x^2(2x - 1)^3\]

We can factor out common terms for further simplification:

\[\frac{dy}{dx} = (2x - 1)^3 \left[ 2x(2x - 1) + 24 + 8x^2 \right]\]

Now, simplifying the expression inside the brackets:

\[2x(2x - 1) = 4x^2 - 2x\]

Thus, we have:

\[\frac{dy}{dx} = (2x - 1)^3 \left[ 4x^2 - 2x + 24 + 8x^2 \right]\]

Combining like terms gives:

\[\frac{dy}{dx} = (2x - 1)^3 (12x^2 - 2x + 24)\]

This is the final expression for the derivative \(\frac{dy}{dx}\). Remember, while the steps may seem complex, staying organized and methodical is key to successfully applying the product and chain rules in calculus.

To find the derivative of the function \( y = \frac{(2x - 1)^4}{3 + x^2} \), we will apply both the quotient rule and the chain rule. The quotient rule states that for a function in the form of \( \frac{u}{v} \), the derivative is given by:

\[\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]

In this case, let \( u = (2x - 1)^4 \) and \( v = 3 + x^2 \). We will first need to find the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) using the chain rule.

For \( u = (2x - 1)^4 \), we apply the chain rule:

\[\frac{du}{dx} = 4(2x - 1)^3 \cdot \frac{d}{dx}(2x - 1) = 4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3\]

For \( v = 3 + x^2 \), the derivative is straightforward:

\[\frac{dv}{dx} = 2x\]

Now, substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula, we have:

\[\frac{dy}{dx} = \frac{(3 + x^2)(8(2x - 1)^3) - (2x - 1)^4(2x)}{(3 + x^2)^2}\]

Next, we simplify the expression. The denominator becomes \( (3 + x^2)^2 \). In the numerator, we can distribute and combine like terms:

\[= \frac{8(3 + x^2)(2x - 1)^3 - 2x(2x - 1)^4}{(3 + x^2)^2}\]

To further simplify, we can factor out common terms. Notice that \( (2x - 1)^3 \) is a common factor:

\[= \frac{(2x - 1)^3 \left[ 8(3 + x^2)(2x - 1) - 2x(2x - 1) \right]}{(3 + x^2)^2}\]

Now, simplifying the expression inside the brackets:

\[= 8(3 + x^2)(2x - 1) - 2x(2x - 1) = 16x - 8 + 8x^3 - 2x(2x - 1)\]

After distributing \( -2x(2x - 1) \), we get:

\[= 16x - 8 + 8x^3 - 4x^2 + 2x = 8x^3 - 4x^2 + 18x - 8\]

Thus, the final derivative is:

\[\frac{dy}{dx} = \frac{(2x - 1)^3 (8x^3 - 4x^2 + 18x - 8)}{(3 + x^2)^2}\]

In conclusion, when finding the derivative of a function that involves both the quotient and chain rules, it is essential to carefully apply each rule step by step, ensuring that all derivatives and simplifications are accurately performed. This process can be complex, but with practice, it becomes more manageable.

To find the derivatives of the functions \( f(x) = \sin^5(x) \) and \( g(x) = \sin(x^5) \), we can apply the chain rule, which is essential for differentiating composite functions.

Starting with the first function, \( f(x) = \sin^5(x) \), we can rewrite it as \( f(x) = (\sin(x))^5 \). Here, the outer function is \( u^5 \) where \( u = \sin(x) \). According to the chain rule, the derivative \( f'(x) \) is given by:

\[f'(x) = 5(\sin(x))^4 \cdot \cos(x)\]

In this expression, we first differentiate the outer function, bringing down the exponent 5 and reducing it by one, while keeping the inner function \( \sin(x) \) intact. Then, we multiply by the derivative of the inner function, which is \( \cos(x) \). Thus, the final derivative for the first function is:

\[f'(x) = 5 \cos(x) \sin^4(x)\]

Next, we consider the second function, \( g(x) = \sin(x^5) \). Here, the outer function is \( \sin(u) \) where \( u = x^5 \). Again, applying the chain rule, we find the derivative \( g'(x) \) as follows:

\[g'(x) = \cos(x^5) \cdot \frac{d}{dx}(x^5)\]

To differentiate \( x^5 \), we use the power rule, which gives us \( 5x^4 \). Therefore, we can express the derivative of the second function as:

\[g'(x) = \cos(x^5) \cdot 5x^4\]

Rearranging this for conventional presentation, we have:

\[g'(x) = 5x^4 \cos(x^5)\]

In summary, the derivatives of the two functions are:

\[f'(x) = 5 \cos(x) \sin^4(x)\]

and

\[g'(x) = 5x^4 \cos(x^5)\]

Practicing these techniques will enhance your comfort with the chain rule, especially when dealing with trigonometric functions.

To find the derivative of a composite function, the chain rule is an essential tool that allows us to differentiate functions that are nested within one another. This process involves working from the outside function to the innermost function, applying the derivative rules at each level. Let's explore this concept through an example.

Consider the function \( f(x) = \sin^4(3x^2) \). This function can be viewed as a composition of three functions: the innermost function \( g(x) = 3x^2 \), the middle function \( h(x) = \sin(g(x)) \), and the outer function \( k(x) = h(x)^4 \). To simplify our differentiation process, we can rewrite the function as \( f(x) = (\sin(3x^2))^4 \), making it clearer how to apply the chain rule.

We start by differentiating the outermost function. Using the power rule, the derivative of \( k(x) = x^n \) is given by \( k'(x) = n \cdot x^{n-1} \). Thus, the derivative of \( f(x) \) with respect to \( x \) begins as:

\[f'(x) = 4(\sin(3x^2))^3 \cdot \frac{d}{dx}[\sin(3x^2)]\]

Next, we differentiate the middle function \( h(x) = \sin(g(x)) \). The derivative of \( \sin(u) \) is \( \cos(u) \), so we multiply by \( \cos(3x^2) \), treating \( 3x^2 \) as a variable:

\[f'(x) = 4(\sin(3x^2))^3 \cdot \cos(3x^2) \cdot \frac{d}{dx}[3x^2]\]

Now, we differentiate the innermost function \( g(x) = 3x^2 \). The derivative is straightforward:

\[\frac{d}{dx}[3x^2] = 6x\]

Combining all these derivatives, we arrive at the full derivative:

\[f'(x) = 4(\sin(3x^2))^3 \cdot \cos(3x^2) \cdot 6x\]

We can simplify this expression by multiplying the constants:

\[f'(x) = 24x(\sin(3x^2))^3 \cdot \cos(3x^2)\]

This final expression represents the derivative of the original function, demonstrating how the chain rule can be effectively applied multiple times to differentiate composite functions. Understanding this process is crucial for tackling more complex derivatives in calculus.

To find the derivative of the function \( f(x) = \tan(3 - \sin(2x)) \), we will apply the chain rule multiple times, as this function is composed of several nested functions. The outermost function is the tangent function, and the argument of the tangent is \( 3 - \sin(2x) \). Within this argument, we have the sine function, which itself contains the innermost function \( 2x \).

We start by applying the chain rule from the outside in. The derivative of the tangent function is given by:

Next, we need to differentiate the inner function \( 3 - \sin(2x) \). The derivative of a constant (3) is 0, and the derivative of \( -\sin(2x) \) is:

Now, we combine these results using the chain rule. The derivative \( f'(x) \) is then:

To simplify, we can rearrange the terms. The final expression for the derivative is:

This expression represents the rate of change of the function \( f(x) \) with respect to \( x \), incorporating the necessary derivatives from the chain rule.