Problem 45 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of the function. \(g(t)=\frac{\sqrt{t+1}}{\sqrt{t^{2}+1}}\)

Step-by-Step Solution

Answer

The derivative of the given function is \[g'(t) = \frac{\sqrt{t^2 + 1} - t\sqrt{t + 1}}{(t^2 + 1)\sqrt{t+1}}\]

Step 1: Identify the functions u(t) and v(t)

In our case, the given function is in the form \(\frac{u(t)}{v(t)}\), where \(u(t) = \sqrt{t + 1}\) and \(v(t) = \sqrt{t^2 + 1}\).

Step 2: Compute the derivatives of u(t) and v(t) with respect to t

For \(u(t) = \sqrt{t + 1}\), we can rewrite it as \((t+1)^{1/2}\). Using the chain rule, we get \(\frac{du}{dt} = \frac{1}{2}(t+1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{t+1}}\). For \(v(t) = \sqrt{t^2 + 1}\), we can rewrite it as \((t^2+1)^{1/2}\). Using the chain rule, we get \(\frac{dv}{dt} = \frac{1}{2}(t^2+1)^{-1/2} \cdot 2t = \frac{t}{\sqrt{t^2+1}}\).

Step 3: Apply the quotient rule formula

Now that we have found \(\frac{du}{dt}\) and \(\frac{dv}{dt}\), we will apply the quotient rule formula to find the derivative of \(\frac{u(t)}{v(t)}\), which is \[\frac{d}{dt} g(t) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2}\] Substituting the values of \(u(t)\), \(v(t)\), \(\frac{du}{dt}\), and \(\frac{dv}{dt}\), we get \[g'(t) = \frac{\sqrt{t^2 + 1}\cdot \frac{1}{2\sqrt{t+1}} - \sqrt{t + 1}\cdot \frac{t}{\sqrt{t^2 + 1}}}{(\sqrt{t^2 + 1})^2}\]

Step 4: Simplify the result

Now, we simplify the expression for the derivative: \[g'(t) = \frac{\frac{\sqrt{t^2 + 1}}{2\sqrt{t+1}} - \frac{t\sqrt{t + 1}}{\sqrt{t^2 + 1}}}{t^2 + 1}\] Multiply the numerator and denominator by \(2\sqrt{t+1}\): \[g'(t) = \frac{2(\sqrt{t^2 + 1}) - 2t\sqrt{t + 1}}{2(t^2 + 1)\sqrt{t+1}}\] Factor out 2 from the numerator: \[g'(t) = \frac{2(\sqrt{t^2 + 1} - t\sqrt{t + 1})}{2(t^2 + 1)\sqrt{t+1}}\] Cancel the common factor 2 from both the numerator and the denominator: \[g'(t) = \frac{\sqrt{t^2 + 1} - t\sqrt{t + 1}}{(t^2 + 1)\sqrt{t+1}}\] The derivative of the given function is \[\boxed{g'(t) = \frac{\sqrt{t^2 + 1} - t\sqrt{t + 1}}{(t^2 + 1)\sqrt{t+1}}}\]

Key Concepts

Chain Rule

The chain rule in calculus is a method used for finding the derivative of composite functions. When we deal with a function within another function, like the square root of an expression, the chain rule allows us to differentiate this composite by taking the derivative of the outside function and then multiplying it by the derivative of the inside function.

For instance, in the exercise provided, to find the derivative of \( u(t) = \(\sqrt{t + 1}\) \), we see \( u(t) \) as the outside function \( \sqrt{} \) applied to the inside function \( t+1 \). By applying the chain rule, we first differentiate the outside function, treating the inside function as a constant, and then multiply by the derivative of the inside function. This gives us \( \frac{du}{dt} = \frac{1}{2}(t+1)^{-1/2} \), simplifying the process of finding derivatives of more complex expressions.

Quotient Rule

The quotient rule is another crucial concept in calculus, particularly when we are tasked with differentiating ratios of functions. This rule states that the derivative of a quotient \( \frac{u(t)}{v(t)} \) is given by \( \frac{v(t)\frac{du}{dt} - u(t)\frac{dv}{dt}}{v(t)^2} \). In essence, it involves differentiating the numerator and denominator separately and then combining the results as specified.

In our exercise, \( g(t) \) is a quotient of two functions involving square roots. After we find the derivatives of the numerator \( u(t) \) and the denominator \( v(t) \) using the chain rule, the quotient rule allows us to put these together to calculate \( g'(t) \).

Simplifying Derivatives

Simplifying derivatives means reducing the derivative expression to its simplest form, making it more comprehensible and easier to evaluate or analyze. In the given exercise, this involves algebraic manipulation such as canceling out common factors and combining like terms. For instance, after applying the quotient rule, we multiplied both the numerator and the denominator by \( 2\sqrt{t+1} \) to get rid of the complex square root in the denominator.

Algebraic simplification can often reveal the nature of the derivative more clearly and is a valuable step in ensuring that the solutions are presented in their most reduced form. Nonetheless, simplification should be conducted cautiously to avoid introducing errors into the derivative expression.

Calculus for Managerial, Life, and Social Sciences

Calculus for managerial, life, and social sciences is a tailored version of calculus that focuses on applications relevant to these fields. Problems, like the one we've worked through, develop the analytical skills needed to model and solve real-world problems. These can range from rates of change in business profits to the spread of a disease in a population.

Understanding calculus concepts, such as the chain and quotient rules, can empower students in these fields to interpret and solve dynamic problems. For example, the derivative calculated in our exercise could represent a rate of change in an economic model or a biological phenomenon, illustrating the value of calculus beyond the realm of pure mathematics.