Math Problem Statement

Buktikan turunan dari f(x) = sec x adalah f'(x) = sec x tan x.

Solution

The question asks to prove that the derivative of f(x)=secxf(x) = \sec x is f(x)=secxtanxf'(x) = \sec x \tan x.

Solution:

Given: f(x)=secxf(x) = \sec x

To find f(x)f'(x), we can use the derivative of secx\sec x with respect to xx.

Derivative of secx\sec x:

We know: ddx(secx)=secxtanx\frac{d}{dx} (\sec x) = \sec x \tan x

Proof:

  1. Rewrite secx\sec x as 1cosx\frac{1}{\cos x}.

    So, f(x)=secx=1cosxf(x) = \sec x = \frac{1}{\cos x}

  2. Use the quotient rule for differentiation: If h(x)=u(x)v(x)h(x) = \frac{u(x)}{v(x)}, then h(x)=u(x)v(x)u(x)v(x)[v(x)]2h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}

    Here, u(x)=1u(x) = 1 and v(x)=cosxv(x) = \cos x.

  3. Compute u(x)u'(x) and v(x)v'(x): u(x)=0u'(x) = 0 v(x)=sinxv'(x) = -\sin x

  4. Substitute into the quotient rule: f(x)=0cosx1(sinx)(cosx)2f'(x) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{(\cos x)^2} f(x)=sinxcos2xf'(x) = \frac{\sin x}{\cos^2 x}

  5. Rewrite sinxcos2x\frac{\sin x}{\cos^2 x} using trigonometric identities: f(x)=secxtanxf'(x) = \sec x \tan x

Thus, we have shown that f(x)=secxtanxf'(x) = \sec x \tan x

Would you like more details or have any questions?

Related Questions:

  1. What is the derivative of tanx\tan x?
  2. How do you prove the derivative of cscx\csc x?
  3. What is the derivative of cotx\cot x?
  4. How is the chain rule applied in trigonometric differentiation?
  5. What is the second derivative of f(x)=secxf(x) = \sec x?

Tip:

Memorizing basic trigonometric derivatives, like those of sinx\sin x, cosx\cos x, tanx\tan x, and secx\sec x, will make solving these kinds of problems faster.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Trigonometric Functions

Formulas

Derivative of sec x is sec x tan x
Quotient Rule: h'(x) = (u'(x)v(x) - u(x)v'(x)) / [v(x)]^2

Theorems

Quotient Rule
Trigonometric Identities

Suitable Grade Level

Grades 11-12