Question

Given \(x=\ln \left(u^{2}-v^{2}\right), u=t^{2}, v=\cos t,\) find \(d x / d t.\)

Short answer

\( \frac{dx}{dt} = \frac{4t^3 + 2 \cos t \sin t}{t^4 - \cos^2 t} \)

Step-by-step solution

Express the given functions

Start by expressing the functions given in the exercise: 1. Let’s note that we have a function of two variables: \( x = \ln(u^2 - v^2) \) 2. We are also given that \( u = t^2 \) and \( v = \cos t \)

Compute the derivatives of u and v with respect to t

Differentiate both u and v with respect to t: 1. \( \frac{du}{dt} = \frac{d}{dt}(t^2) = 2t \) 2. \( \frac{dv}{dt} = \frac{d}{dt}( \cos t ) = -\sin t \)

Apply the chain rule to find \( \frac{dx}{dt} \)

Using the chain rule, we write the derivative of x with respect to t as: 1. From \( x = \ln(u^2 - v^2) \), use the formula for the derivative of a logarithm: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \frac{dx}{dt} \) 2. Therefore, \( \frac{dx}{dt} = \frac{1}{u^2 - v^2} \frac{d}{dt}(u^2 - v^2) \)

Differentiate \(u^2 - v^2\) with respect to t

Differentiate \(u^2 - v^2\) with respect to t: 1. First we find: \( \frac{d}{dt}(u^2) = \frac{d}{dt}(t^4) = 4t^3 \) 2. Then: \( \frac{d}{dt}(v^2) = \frac{d}{dt}(\cos^2 t) = 2 \cos t (-\sin t) = -2 \cos t \sin t \) 3. Thus, \( \frac{d}{dt}(u^2 - v^2) = 4t^3 - (-2 \cos t \sin t) = 4t^3 + 2 \cos t \sin t \)

Combine the results

Combine all previous results to find \( \frac{dx}{dt} \): 1. \( \frac{dx}{dt} = \frac{1}{u^2 - v^2} \cdot (4t^3 + 2 \cos t \sin t) \) 2. Substitute back \( u = t^2 \) and \( v = \cos t \): \( \frac{dx}{dt} = \frac{4t^3 + 2 \cos t \sin t}{t^4 - \cos^2 t} \)

Key concepts

chain rule

Understanding the chain rule is crucial for differentiating composite functions. The chain rule allows us to take the derivative of a function within a function. Suppose we have a function composed of two functions, say, function h(x) which is f(g(x)). The chain rule states that the derivative of h(x) with respect to x is given by the product of the derivative of f with respect to g and the derivative of g with respect to x. In mathematical terms, if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \times g'(x) \). In the exercise, we applied the chain rule to find \( \frac{dx}{dt} \) for the given expressions. This involves differentiating \( x = \ln(u^2 - v^2) \) and then utilizing the derivatives of u and v with respect to t.

It's essential to break it down step-by-step when using the chain rule:

• Compute the inner derivatives first (the u and v functions).
• Then use these in the derivative of the composed function.

logarithmic functions

Logarithmic functions, specifically the natural logarithm (ln), appear frequently in calculus. The natural logarithm is the inverse of the exponential function. When differentiating logarithmic functions, the key rule to remember is: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \frac{dx}{dt} \). This rule is particularly useful when the argument of the logarithm is a function itself, making the use of the chain rule necessary.

In the given exercise, the expression \( x = \ln(u^2 - v^2) \) requires applying this differentiation rule to logarithms. The steps involve:

• First, differentiating the logarithmic function by applying the rule above.
• Then, differentiating the inside function \( u^2 - v^2 \).

Comprehending these steps simplifies complex logarithmic differentiation and makes it manageable.

differentiation with respect to a parameter

Differentiation with respect to a parameter involves finding the derivative of a function in terms of a given variable or parameter. In our exercise, we are asked to find \( \frac{dx}{dt} \), meaning we need to differentiate x with respect to the parameter t.

The process involves:

• Identifying the functions that depend on t (in this case u and v).
• Differentiating each function u and v with respect to t.
• Applying the chain rule to link these derivatives back to the original function x.

This approach allows us to understand how changes in t affect the overall function x, revealing the relationship between these variables.

partial derivatives

Partial derivatives are derivatives of functions with multiple variables, where we differentiate with respect to one variable while keeping the others constant. This is central in multivariable calculus. If we have a function f(x, y), the partial derivative with respect to x is denoted as \( \frac{\partial f}{\partial x} \), and it involves treating y as a constant.

In the given exercise, partial derivatives come into play when differentiating expressions involving more than one variable (like u and v).

Here’s how to approach partial derivatives:

• Examine the function and identify the dependent and independent variables.
• Treat all other variables as constants while differentiating with respect to the specified variable.

Mastering partial derivatives helps in understanding how each variable individually influences the function, critical for problems involving functions of multiple variables.