Question

In Exercises \(33-38,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1$$

Short answer

The value of \((f \circ g)^{\prime}\) at \(x = 1\) is 2.5.

Step-by-step solution

Find the Derivative of the Inner Function

The first thing we need to do is find the derivative of the inner function \( g(x)=\sqrt{x} \). Using the power rule for derivatives \( g^{\prime}(x) = 0.5 x^{-1/2} \), where we move the exponent in front and subtract 1 from it.

Find the Derivative of the Outer Function

Next, find the derivative of the outer function \( f(u)=u^{5}+1 \). Consider \( u \) as \( x \) temporarily and apply the power rule, which yields \( f'(u) = 5u^4 \). We don't include the derivative of the constant since it's zero.

Apply the Chain Rule

To find the derivative of the composite function \((f \circ g)(x)\), we apply the chain rule. The chain rule states that \( (f \circ g)^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x) \). Substituting the values found in steps 1 and 2, we get the derivative of the composite function as \( (f \circ g)^{\prime}(x)=5g(x)^4 \cdot 0.5 x^{-1/2} = 2.5x^{-1/2} \cdot g(x) ^4 \).

Evaluate the Derivative at x=1

Finally, we substitute the given value of \( x \) into our derivative to compute the final result. Putting \( x=1\) into \(2.5x^{-1/2} \cdot g(x)^4 \), we get \(2.5 \cdot 1^{-1/2} \cdot (\sqrt{1})^4 = 2.5 \cdot 1 \cdot 1 = 2.5 \)

Key concepts

Composite Function Derivative

Understanding how to differentiate a composite function is an essential skill when studying calculus. A composite function is created when one function is applied within another function; in other words, you have a function inside of another function. For example, if we have two functions, f(u) and g(x), the composite function is written as (f \textcircled g)(x) = f(g(x)). The derivative of this composite function isn't as straightforward as taking the derivatives separately but involves an application of the chain rule. The chain rule allows us to differentiate a composite function by multiplying the derivative of the outer function by the derivative of the inner function, after the inner function has been substituted into the outer function.This can be a complex concept to grasp, so let’s illustrate it through an example. If we have a function 'f' defined by f(u) and an inner function 'g' by g(x), the derivative of the composite function at a point x would be determined by evaluating the derivative of f at g(x) and then multiplying that by the derivative of g at x. For the specific exercise given, where f(u)=u^5+1 and g(x)=\(\text{sqrt}){x}\), the composite function could be expressed as (f \textcircled g)(x) = (g(x))^5 + 1.

Power Rule for Derivatives

When it comes to taking derivatives, the power rule is one of the most frequently used techniques. The power rule states that for any function of the form f(x) = x^n, the derivative f'(x) is nx^{n-1}. It simplifies the process significantly when dealing with polynomial functions or any term that can be expressed as a power of x.To put this into practice, consider the function f(u) = u^5 from our example. Using the power rule, the derivative f'(u) would therefore be 5u^4. Notice that constants, like the '+1' in our example, disappear in the derivative, as their rate of change is zero.

Chain Rule Application

Applying the chain rule involves a two-step process, where you first find the derivatives of the inner and outer functions separately and then multiply them together. It's widely used for more complex derivatives involving composite functions, as seen in the exercise at hand.For instance, if h(x) = (f \textcircled g)(x), then h'(x) can be found using the chain rule: h'(x) = f'(g(x)) \(\times\) g'(x). This method shows why understanding both the power rule for finding derivatives and how to determine derivatives of individual functions is crucial before one can approach composite functions and effectively use the chain rule. The solution to our example demonstrates this beautifully by breaking down the process into manageable steps.

Derivative of Inner and Outer Functions

When dealing with the derivative of composite functions, distinguishing between the 'inner' and 'outer' functions is pivotal. The 'outer' function is the one that takes the 'inner' function as its argument. In the context of our exercise, f(u) represents the outer function, and g(x), the inner function.By convention, we first differentiate the outer function as if the inner function were a regular variable, and then multiply it by the derivative of the inner function. For our specific example, the derivative of the outer function f(u)=u^5 + 1 is f'(u)=5u^4, and the derivative of the inner function g(x)=\(\text{sqrt}){x}\) or x^{1/2} is g'(x)=1/(2\(\text{sqrt}){x}\)). These will then be used to find the derivative of the composite function by following the chain rule, producing a final expression that represents how rapidly the composite function changes in relation to x.