Question

In Exercises \(1-8,\) use the given substitution and the Chain Rule to find \(d y / d x\) $$y=\cos (\sqrt{3} x), u=\sqrt{3 x}$$

Short answer

The derivative of \(y = \cos(\sqrt{3x})\) with respect to \(x\) is \(dy/dx = -\sin(\sqrt{3x}) * \frac{3}{2 \sqrt{x}}\).

Step-by-step solution

Apply the substitution

Use the given substitution to replace \(\sqrt{3x}\) in \(y = \cos(\sqrt{3x})\). This leads to a new function, \(y = \cos(u)\).

Find derivative of the outer function

Calculate the derivative of the function \(y = \cos(u)\) with respect to \(u\). The derivative of \(\cos(u)\) is -\(\sin(u)\), hence \(dy/du = -\sin(u)\).

Find derivative of the inner function

Now, find the derivative of \(u = \sqrt{3x}\) with respect to \(x\). This becomes \(du/dx = 3/(2\sqrt{x})\).

Apply Chain Rule to find dy/dx

Using the Chain Rule, multiply the derivative of the outer function by the derivative of the inner function to find the derivative of \(y\) with respect to \(x\), which is \(dy/dx = (dy/du) * (du/dx)\). This results in \(dy/dx = -\sin(u) * \frac{3}{2 \sqrt{x}}\).

Substitute back the value of u

Substitute \(u = \sqrt{3x}\) into the result from the previous step, to get the derivative of the original function in terms of \(x\). So, \(dy/dx = -\sin(\sqrt{3x}) * \frac{3}{2 \sqrt{x}}\).

Key concepts

Derivative of Trigonometric Functions

Understanding how to find the derivative of trigonometric functions is a crucial part of calculus. Trigonometric functions include sine (\text{sin}), cosine (\text{cos}), and tangent (\text{tan}), among others. These functions often appear in a variety of physics and engineering problems, as well as in calculus exercises.

For example, the derivative of the cosine function with respect to a variable, say, \text{u}, is given by \( \frac{d}{du} \cos(u) = -\sin(u) \). Similarly, the derivative of sine is \( \frac{d}{du} \sin(u) = \cos(u) \), and for tangent, it’s \( \frac{d}{du} \tan(u) = \sec^2(u) \). These derivatives can be used later with the help of rules like the Chain Rule to find derivatives of more complex functions involving trigonometric terms.

To master the differentiation of trigonometric functions, it's recommended to memorize these core derivatives, as they are often the building blocks for more advanced calculations.

Implicit Differentiation

Implicit differentiation is a technique used when facing equations where the dependent variable, commonly \text{y}, and the independent variable, typically \text{x}, are intermixed. Rather than explicitly solving for \text{y} in terms of \text{x}, you differentiate both sides of the equation with respect to \text{x}, applying the Chain Rule as necessary.

For example, if you have an equation like \( x^2 + y^2 = 1 \), it's easier to differentiate with respect to \text{x} directly than to solve for \text{y} first. By treating \text{y} as an implicit function of \text{x}, you can take the derivative of \text{y} with respect to \text{x} as if \text{y} were an explicit function of \text{x}. This often results in derivatives that include \( \frac{dy}{dx} \) on both sides of the equation, which you then solve algebraically to find \( \frac{dy}{dx} \).

Substitution Method in Calculus

The substitution method is widely used to simplify the process of finding derivatives and is especially powerful when dealing with composition of functions, as seen in the Chain Rule. The basic idea is to replace a complicated or cumbersome expression with a single variable, making differentiation more manageable.

In the context of our exercise, we replaced \( \sqrt{3x} \) with \( u \), transforming the function into a simpler form, which is then differentiated. After finding the derivative with respect to the new variable, you substitute back the original expression. This method streamlines the differentiation process, but more importantly, it sets the stage for applying the Chain Rule by isolating the 'inner function' and the 'outer function.'

Finding Derivatives

Finding derivatives, or differentiation, is the process of calculating the rate at which a function is changing at any given point. It's a fundamental operation in calculus that gets applied across various scientific fields. There are several rules and techniques to find derivatives, such as the Power Rule, Product Rule, Quotient Rule, and the Chain Rule, among others.

In practical terms, finding derivatives involves understanding these rules and knowing when and how to apply them to different types of functions—whether they're polynomials, trigonometric, exponential, or implicit functions. When it comes to applying these rules, practice is key. Work through problems step by step and always keep in mind the function type and the most appropriate rule or combination of rules to use. This will allow you to efficiently tackle a vast array of problems in calculus.