Chapter 2: Problem 36
Find \(d y / d x\) $$ y=\left(x^{2}+x\right)^{5} \sin ^{8} x $$
Short Answer
Expert verified
\( \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \sin^8 x + 8(x^2 + x)^5 \sin^7 x \cos x \).
Step by step solution
01
Identify the Rule of Differentiation
The function given is a product of two functions, \( (x^2 + x)^5 \) and \( \sin^8x \). Therefore, we will use the product rule to differentiate it.
02
Apply the Product Rule
The product rule states that if you have a function \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \). Here \( u = (x^2 + x)^5 \) and \( v = \sin^8x \). We'll differentiate each part separately.
03
Differentiate \( u = (x^2 + x)^5 \) Using the Chain Rule
Using the chain rule, let \( g(x) = x^2 + x \) so that \( u = g(x)^5 \). Then, \( u' = 5g(x)^4 g'(x) \). Since \( g(x) = x^2 + x \), we have \( g'(x) = 2x + 1 \). So, \( u' = 5(x^2 + x)^4 (2x + 1) \).
04
Differentiate \( v = \sin^8x \) Using the Chain Rule
Again using the chain rule, let \( h(x) = \sin x \) so that \( v = h(x)^8 \). Then, \( v' = 8 h(x)^7 h'(x) \). Since \( h'(x) = \cos x \), \( v' = 8 \sin^7 x \cos x \).
05
Compute \( \frac{dy}{dx} \) Using the Product Rule Results
Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule: \[ \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \cdot \sin^8 x + (x^2 + x)^5 \cdot 8 \sin^7 x \cos x \].
06
Simplify the Expression
Simplify the expression to make it more manageable if needed. As is, \( \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \sin^8 x + 8(x^2 + x)^5 \sin^7 x \cos x \) is already quite concise.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
To find the derivative of a product of two functions, we use the product rule. The product rule is a powerful tool in calculus, allowing us to deal with functions that are multiplied together. If you have a function defined as a product of two separate functions, say \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} \) is given by the formula:
- \( \frac{dy}{dx} = u'v + uv' \)
Chain Rule
When functions are nested inside each other, like \((x^2 + x)^5\), we use the chain rule to differentiate them. The chain rule states that if a function \( y \) is composed of two functions \( g(x) \) and \( f(u) \), where \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) is given by:
- \( \frac{dy}{dx} = f'(u) \cdot g'(x) \)
Derivative of Trigonometric Functions
Differentiating trigonometric functions is a common task in calculus. These functions can be part of complex expressions, just like \( \sin^8 x \). Here, the chain rule aids in finding their derivative. To differentiate \( \sin^8x \), consider it as \( (\sin x)^8 \). Here, \( h(x) = \sin x \) serves as an inner function. The derivative involves:
- First differentiate the outer function: \( 8(\sin x)^7 \)
- Then multiply by the derivative of the inner function: \( \cos x \)