Math Doubts

$\dfrac{d}{dx}{\,\sin{(ax)}}$ rule

Formula

$\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a\cos{(ax)}$

Introduction

Let $a$ and $x$ be a constant and a variable respectively but the variable $x$ represents an angle in this case. The product of $a$ and $x$ is $ax$, which represents a multiple angle in mathematical form. The sine of a multiple angle $ax$ is written as $\sin{(ax)}$ mathematically.

The derivative of sine of a multiple angle $ax$ with respect to $x$ is written in the following mathematical form in calculus.

$\dfrac{d}{dx}{\,\sin{(ax)}}$

The derivative of sine of multiple angle $ax$ with respect to $x$ is equal to the product of multiple constant $a$ and the cosine of multiple angle $ax$.

$\implies$ $\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a \times \cos{(ax)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\sin{(ax)}}$ $\,=\,$ $a\cos{(ax)}$

It is called the derivative rule for the sine of a multiple angle.

Use

It is used as a formula to find the derivative of a sine function in which multiple or submultiple angle is involved.

Example

Find $\dfrac{d}{dx}{\,\sin{(2x)}}$

In this example, $a \,=\, 2$, so, substitute it in the derivative of sine of multiple angle formula to find the derivative of $\sin{(2x)}$ with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,\sin{(2x)}}$ $\,=\,$ $2 \times \cos{(2x)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\sin{(2x)}}$ $\,=\,$ $2\cos{(2x)}$

Proof

Learn how to prove the differentiation formula for finding the derivative of the sine of a multiple angle with respect to a variable.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved