Problem 180 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for the given functions. $$ y=\sin x \tan x $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \sin x + \sin x \sec^2 x \).

Step 1: Understand the Product Rule

The function given is a product of two sub-functions, \( y = u(x) \cdot v(x) \) where \( u(x) = \sin x \) and \( v(x) = \tan x \). To find the derivative \( \frac{d y}{d x} \), the product rule is applied: \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x). \] This rule states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second.

Step 2: Compute the Derivatives of the Sub-functions

Calculate the derivatives of \( \sin x \) and \( \tan x \): - The derivative of \( \sin x \) is \( \cos x \). Thus, \( u'(x) = \cos x \).- The derivative of \( \tan x \) is \( \sec^2 x \). Thus, \( v'(x) = \sec^2 x \).

Step 3: Apply the Product Rule

Substitute \( u(x) \), \( v(x) \), and their respective derivatives into the product rule formula:\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x), \]which becomes:\[ \frac{dy}{dx} = (\cos x)(\tan x) + (\sin x)(\sec^2 x). \]

Step 4: Simplify the Expression

Use trigonometric identities to simplify the expression, where possible:- Recall \( \tan x = \frac{\sin x}{\cos x} \), so \( \cos x \tan x = \sin x \).Thus, the expression becomes:\[ \frac{dy}{dx} = \sin x + \sin x \sec^2 x. \]No further simplification is possible using basic identities, so this is the final derivative.

Key Concepts

Trigonometric Derivatives

Trigonometric functions are fundamental in calculus, and knowing their derivatives is crucial for solving various problems. In the case of \( y = \sin x \tan x \), we deal with two primary trigonometric functions: sine and tangent. To successfully differentiate such expressions, one must know their respective derivatives.

The derivative of \( \sin x \) is \( \cos x \), a basic derivative you should remember. It stems from the chain rule applied to sine and the relation between sine and cosine.

Tangent, another common trig function, has a derivative of \( \sec^2 x \). This derivative originates from the identity \( \tan x = \frac{\sin x}{\cos x} \) and employing the quotient rule. Understanding these derivatives helps in tackling problems where trigonometric functions are involved, especially in complex expressions.

Calculus Derivatives

When dealing with calculus derivatives, it's important to recognize different rules and techniques available, like the product rule for this exercise. The product rule is fundamental when differentiating products of functions.

For a product \( u(x) \cdot v(x) \), the product rule states:

• Take the derivative of the first function, \( u'(x) \), multiply by the second function, \( v(x) \).
• Add the first function, \( u(x) \), multiplied by the derivative of the second function, \( v'(x) \).

For \( y = \sin x \tan x \), applying the product rule combines these into one expression: \( \cos x \tan x + \sin x \sec^2 x \). This systematic approach simplifies working with more complicated functions, making them easier to understand and solve.

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions involves using identities to make them more manageable or to reveal hidden patterns. In the context of this problem, simplifying is the final step after using derivatives.

To simplify an expression like \( \cos x \tan x + \sin x \sec^2 x \), you can employ identities like \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec x = \frac{1}{\cos x} \). Here, \( \cos x \tan x \) simplifies to \( \sin x \), thanks to replacing \( \tan x \) with its equivalent \( \frac{\sin x}{\cos x} \).

Though it doesn't reduce further in this case, recognizing when to use trigonometric identities can streamline finding solutions and is a valuable tool in derivative problems and integrals alike. By simplifying, you make expressions clearer and often open the way to discover deeper insights into the problem at hand.