Derivative of Sin Square x by First Principle

Formula to find derivative of a function f(x) by first principle :

This is also called as limit definition of the derivative.

Derivative of sin2x using first principle :

Let

f(x) = sin2x

Trigonometric Identities :

sin(A + B) = sinAcosB + cosAsinB

sin(A - B) = sinAcosB - cosAsinB

sin(A + B)  sin(A - B) :

= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)

= (sinAcosB)2 - (cosAsinB)2

= sin2Acos2B - cos2Asin2B

= sin2A(1 - sin2B) - (1 - sin2A)sin2B

= sin2A - sin2Asin2B - sin2B + sin2Asin2B

= sin2A - sin2B

Therefore,

sin2A - sin2B = sin(A + B)  sin(A - B)

From standard results of limits,


Therefore,

(sin2x)' = sin(2x)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Adaptive Learning Platforms

    May 26, 23 12:27 PM

    adaptivelearning1
    Adaptive Learning Platforms: Personalized Mathematics Instruction with Technology

    Read More

  2. Simplifying Expressions with Rational Exponents Worksheet

    May 21, 23 07:40 PM

    tutoring.png
    Simplifying Expressions with Rational Exponents Worksheet

    Read More

  3. Simplifying Rational Expressions Worksheet

    May 20, 23 10:53 PM

    tutoring.png
    Simplifying Rational Expressions Worksheet

    Read More