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 \( u = 4x + 5 \). The first step is to differentiate the outer function \( f(u) \). Using the power rule, the derivative of \( u^3 \) is:

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

Next, we substitute back the inside function \( u \) to get:

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

Now, we need to find the derivative of the inside function \( g(x) = 4x + 5 \). The derivative is simply:

\[g'(x) = 4\]

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 inside function as \( g(x) = 3x^2 - x \) and the outside function as \( f(u) = 2u^4 \).

Applying the chain rule, we first differentiate the outer function:

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

Substituting back gives:

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

Next, we differentiate the inside function:

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

Finally, 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 inside and outside functions, applying the power rule, and multiplying by the derivative of the inside function, 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 function consists of two parts being multiplied: \(u = (2x - 1)^4\) and \(v = 3 + x^2\). According to the product rule, the derivative of a product \(uv\) is given by:

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

We will calculate \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) separately. Starting with \(v\), we find:

\[\frac{dv}{dx} = \frac{d(3 + x^2)}{dx} = 0 + 2x = 2x\]

Next, we apply the chain rule to find \(\frac{du}{dx}\). Here, \(u = (2x - 1)^4\) has an inside function \(g(x) = 2x - 1\) and an outside function \(f(g) = g^4\). The chain rule states:

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

Calculating \(f'(g)\) gives us:

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

And for \(g'(x)\), we have:

\[g'(x) = \frac{d(2x - 1)}{dx} = 2\]

Thus, we can combine these results to find \(\frac{du}{dx}\):

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

Now, substituting back into the product rule formula, we have:

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

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

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

And the second term becomes:

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

Combining these, we can factor out common terms. The final expression for the derivative is:

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

After further simplification, we can express the derivative in a more organized manner, ensuring that constants are factored out appropriately. The final result captures the complexity of the original function while adhering to the rules of differentiation.

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 differentiate \( u \) and \( v \) using the chain rule.

For \( u = (2x - 1)^4 \), applying the chain rule involves differentiating the outer function and then the inner function:

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

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

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

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

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

Next, we simplify the expression. The denominator becomes \( (3 + x^2)^2 \). In the numerator, we can factor out common terms:

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

To further simplify, we can factor out \( (2x - 1)^3 \) from the numerator:

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

Now, simplifying the expression inside the brackets:

\[8(3 + x^2) - 2x(2x - 1) = 24 + 8x^2 - 4x^2 + 2x = 24 + 4x^2 + 2x\]

Thus, the derivative can be expressed as:

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

This final expression represents the derivative of the given function, combining the quotient and chain rules effectively. Remember to check each step for accuracy and ensure all terms are accounted for during differentiation.