Find The Derivative Of The Function Defined Below.${ Y = -6x^2 \sin 5x }$ {\frac{dy}{dx} = \}

Mar 9, 2025 by ADMIN 96 views

**Finding the Derivative of a Trigonometric Function**

Introduction

In calculus, the derivative of a function represents the rate of change of the function with respect to one of its variables. In this article, we will focus on finding the derivative of a function that involves a trigonometric function, specifically the sine function. The function we will be working with is defined as:

${ y = -6x^2 \sin 5x }$

Understanding the Function

Before we proceed with finding the derivative, let's take a closer look at the function. The function is a product of two functions: $-6x^2$ and $\sin 5x$ . The first function is a quadratic function, while the second function is a trigonometric function. The derivative of a product of two functions can be found using the product rule, which states that if $y = u(x)v(x)$ , then $y' = u'(x)v(x) + u(x)v'(x)$ .

Applying the Product Rule

To find the derivative of the given function, we will apply the product rule. Let's identify the two functions $u(x)$ and $v(x)$ :

${ u(x) = -6x^2 }$ ${ v(x) = \sin 5x }$

Finding the Derivatives of u(x) and v(x)

To apply the product rule, we need to find the derivatives of $u(x)$ and $v(x)$ . The derivative of $u(x)$ is:

${ u'(x) = \frac{d}{dx} (-6x^2) = -12x }$

The derivative of $v(x)$ is:

${ v'(x) = \frac{d}{dx} (\sin 5x) = 5 \cos 5x }$

Applying the Product Rule

Now that we have found the derivatives of $u(x)$ and $v(x)$ , we can apply the product rule to find the derivative of the given function:

${ y' = u'(x)v(x) + u(x)v'(x) }$ ${ y' = (-12x)(\sin 5x) + (-6x^2)(5 \cos 5x) }$ ${ y' = -12x \sin 5x - 30x^2 \cos 5x }$

Simplifying the Derivative

The derivative we obtained is a bit complicated, so let's simplify it by combining like terms:

${ y' = -12x \sin 5x - 30x^2 \cos 5x }$

Conclusion

In this article, we found the derivative of a function that involves a trigonometric function, specifically the sine function. We applied the product rule to find the derivative, and then simplified the result to obtain the final answer. The derivative of the given function is:

${ \frac{dy}{dx} = -12x \sin 5x - 30x^2 \cos 5x }$

Example Problems

To reinforce your understanding of the product rule, try solving the following example problems:

Find the derivative of the function $y = 2x^3 \cos 2x$ .
Find the derivative of the function $y = x^2 \sin 3x$ .

Practice Problems

To practice finding the derivatives of functions that involve trigonometric functions, try solving the following practice problems:

Find the derivative of the function $y = 3x^4 \sin x$ .
Find the derivative of the function $y = x^3 \cos 2x$ .

Glossary of Terms

Derivative: A measure of how a function changes as its input changes.
Product Rule: A rule for finding the derivative of a product of two functions.
Trigonometric Function: A function that involves the sine, cosine, or tangent of an angle.
Quadratic Function: A function of the form $ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants.

References

[1] Calculus, 3rd edition, by Michael Spivak.
[2] Calculus, 2nd edition, by James Stewart.

Additional Resources

[1] Khan Academy: Calculus.
[2] MIT OpenCourseWare: Calculus.

Introduction

In our previous article, we discussed how to find the derivative of a function that involves a trigonometric function, specifically the sine function. In this article, we will provide a Q&A section to help you better understand the concept and apply it to different problems.

Q: What is the derivative of the function y = sin(x)?

A: The derivative of the function y = sin(x) is y' = cos(x).

Q: How do I find the derivative of a function that involves a trigonometric function?

A: To find the derivative of a function that involves a trigonometric function, you can use the product rule. The product rule states that if y = u(x)v(x), then y' = u'(x)v(x) + u(x)v'(x).

Q: What is the product rule?

A: The product rule is a rule for finding the derivative of a product of two functions. It states that if y = u(x)v(x), then y' = u'(x)v(x) + u(x)v'(x).

Q: How do I apply the product rule to find the derivative of a function that involves a trigonometric function?

A: To apply the product rule, you need to identify the two functions u(x) and v(x), and then find their derivatives. Once you have found the derivatives, you can plug them into the product rule formula to find the derivative of the original function.

Q: What is the derivative of the function y = 2x^3 sin(x)?

A: To find the derivative of the function y = 2x^3 sin(x), you can use the product rule. Let u(x) = 2x^3 and v(x) = sin(x). Then, u'(x) = 6x^2 and v'(x) = cos(x). Plugging these values into the product rule formula, you get:

y' = u'(x)v(x) + u(x)v'(x) y' = (6x^2)(sin(x)) + (2x^3)(cos(x))

Q: What is the derivative of the function y = x^2 cos(2x)?

A: To find the derivative of the function y = x^2 cos(2x), you can use the product rule. Let u(x) = x^2 and v(x) = cos(2x). Then, u'(x) = 2x and v'(x) = -2sin(2x). Plugging these values into the product rule formula, you get:

y' = u'(x)v(x) + u(x)v'(x) y' = (2x)(cos(2x)) + (x^2)(-2sin(2x))

Q: What are some common trigonometric functions and their derivatives?

A: Some common trigonometric functions and their derivatives are:

sin(x) -> cos(x)
cos(x) -> -sin(x)
tan(x) -> sec^2(x)
cot(x) -> -csc^2(x)

Q: How do I find the derivative of a function that involves a trigonometric function with a coefficient?

A: To find the derivative of a function that involves a trigonometric function with a coefficient, you can use the product rule. For example, if you have a function like y = 3x^2 sin(x), you can let u(x) = 3x^2 and v(x) = sin(x). Then, u'(x) = 6x and v'(x) = cos(x). Plugging these values into the product rule formula, you get:

y' = u'(x)v(x) + u(x)v'(x) y' = (6x)(sin(x)) + (3x^2)(cos(x))