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Chapter 29: Problem 10

Find the differential coefficient of \(y=3 x^{2} \sin 2 x\)

Short Answer

Expert verified
The differential coefficient is \(6x(\sin(2x) + x\cos(2x))\).

Step by step solution

01

Identify the Function Type

The given function is a product of two functions: the quadratic function \(u = 3x^2\) and the trigonometric function \(v = \sin(2x)\). We will use the product rule for differentiation.
02

Differentiate Each Component

Differentiate \(u = 3x^2\) with respect to \(x\), which gives \(u' = 6x\). Differentiate \(v = \sin(2x)\) using the chain rule, resulting in \(v' = 2\cos(2x)\).
03

Apply the Product Rule

Use the product rule \((uv)' = u'v + uv'\) to find the derivative of \(y\). Substitute the components: \((3x^2 \sin(2x))' = 6x \sin(2x) + 3x^2 (2\cos(2x))\).
04

Simplify the Expression

Simplify the derivative: \(6x \sin(2x) + 6x^2 \cos(2x)\). Factor out 6x to get \(6x(\sin(2x) + x\cos(2x))\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
In differential calculus, the product rule is essential when differentiating functions that are multiplied together. Typically, when we have two functions, \( u(x) \) and \( v(x) \), their product is \( y = u(x)v(x) \). The product rule states that the derivative of this product is given by:
  • \((uv)' = u'v + uv'\)
This rule is extremely valuable because it simplifies the process of differentiation when dealing with products of functions. For example, in the exercise, we have \( y = 3x^2 \sin(2x) \). Here, \( u(x) = 3x^2 \) and \( v(x) = \sin(2x) \). By differentiating each function separately and applying the product rule, we can find the derivative of the entire expression efficiently.
Chain Rule
The chain rule is crucial for finding the derivative of composite functions. Imagine a function \( y \) that can be expressed as \( y = f(g(x)) \), where \( f \) and \( g \) are two separate functions. The chain rule helps to differentiate \( y \) with respect to \( x \) as follows:
  • \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
This is like a method of unwinding nested functions. To apply this in the original exercise, we look at \( v = \sin(2x) \). Here, \( g(x) = 2x \) and \( f(g) = \sin(g) \). When we differentiate using the chain rule, we find \( v' = 2\cos(2x) \), which allows us to account for both the inner and outer functions in the differentiation process.
Derivative of Trigonometric Functions
Trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) each have well-defined derivatives. These derivatives are essential to solving many problems in calculus. Here's a quick reference list:
  • \( \frac{d}{dx}[\sin(x)] = \cos(x) \)
  • \( \frac{d}{dx}[\cos(x)] = -\sin(x) \)
  • \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \)
In our exercise, the function \( \sin(2x) \) is differentiated to find \( v' \). Using the rule for \( \sin(x) \) along with the chain rule, we derived \( 2\cos(2x) \). This shows how trigonometric derivatives, combined with the chain rule, help in simplifying complex expressions into manageable components.

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Most popular questions from this chapter

Differentiate, with respect to \(x\), $$ y=5 x^{4}+4 x-\frac{1}{2 x^{2}}+\frac{1}{\sqrt{x}}-3 $$

Find the differential coefficient of $$ y=\frac{2}{\left(2 t^{3}-5\right)^{4}} $$

Find the derivatives of (a) \(y=3 \sqrt{x}\) (b) \(y=\frac{5}{\sqrt[3]{x^{4}}}\)

. Determine the gradient of the curve \(y=\frac{5 x}{2 x^{2}+4}\) at the point \(\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right)\)

Determine the differential coefficient of \(y=\sqrt{\left(3 x^{2}+4 x-1\right)}\)
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