Math Problem Statement

(Secx cscx) /csc^2 - sec^2 find derivative

Solution

To find the derivative of the given expression sec(x)csc(x)csc2(x)sec2(x)\frac{\sec(x) \csc(x)}{\csc^2(x)} - \sec^2(x), let's break it down step by step.

Step 1: Simplify the expression

Start by simplifying the first term:

sec(x)csc(x)csc2(x)=sec(x)csc(x)\frac{\sec(x) \csc(x)}{\csc^2(x)} = \frac{\sec(x)}{\csc(x)}

So, the expression becomes:

sec(x)csc(x)sec2(x)\frac{\sec(x)}{\csc(x)} - \sec^2(x)

Now, recall that:

sec(x)=1cos(x),csc(x)=1sin(x)\sec(x) = \frac{1}{\cos(x)}, \quad \csc(x) = \frac{1}{\sin(x)}

Thus:

sec(x)csc(x)=1cos(x)1sin(x)=sin(x)cos(x)=tan(x)\frac{\sec(x)}{\csc(x)} = \frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}} = \frac{\sin(x)}{\cos(x)} = \tan(x)

So the simplified expression is:

tan(x)sec2(x)\tan(x) - \sec^2(x)

Step 2: Find the derivative

Now, let's differentiate the expression with respect to xx:

ddx[tan(x)sec2(x)]\frac{d}{dx} \left[\tan(x) - \sec^2(x)\right]

The derivative of tan(x)\tan(x) is sec2(x)\sec^2(x), and the derivative of sec2(x)-\sec^2(x) is 2sec2(x)tan(x)-2\sec^2(x) \tan(x) (using the chain rule).

So, the derivative is:

sec2(x)2sec2(x)tan(x)\sec^2(x) - 2\sec^2(x)\tan(x)

This is the derivative of the original expression.

Would you like more details on any of these steps? Here are some related questions:

  1. How do you differentiate tan(x)\tan(x) and sec(x)\sec(x)?
  2. Can you simplify the expression further?
  3. What is the chain rule in differentiation?
  4. How does trigonometric identity play a role in differentiation?
  5. Can we find the derivative using an alternative method?

Tip: Always simplify the expression as much as possible before differentiating; this often makes the process easier.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Differentiation

Formulas

Derivative of trigonometric functions: d/dx(tan(x)) = sec^2(x)
Simplification using trigonometric identities: sec(x) = 1/cos(x), csc(x) = 1/sin(x)

Theorems

Chain Rule
Trigonometric Identities

Suitable Grade Level

Grades 11-12, College Level