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Magazine Chapter 23: Problem 19
Find the second derivative of each of the given functions. $$y=2(2-5 x)^{4}$$
Short Answer
Expert verified
The second derivative is \( \frac{d^2y}{dx^2} = 600(2 - 5x)^2. \)
Step by step solution
01
Understand the Function
The given function is in the form of a power of a binomial: \[ y = 2(2 - 5x)^4 \] Our task is to find the second derivative of this function.
02
Find the First Derivative
To find the first derivative, we use the chain rule. Let \[ u = 2 - 5x. \] Then \[ y = 2u^4. \] The derivative of \( y \) with respect to \( u \) is:\[ \frac{dy}{du} = 8u^3. \] Now, differentiate \( u = 2 - 5x \) with respect to \( x \):\[ \frac{du}{dx} = -5. \] Using the chain rule, the first derivative \( \frac{dy}{dx} \) is:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 8(2 - 5x)^3 \cdot (-5) = -40(2 - 5x)^3. \]
03
Find the Second Derivative
Now we differentiate \( \frac{dy}{dx} = -40(2 - 5x)^3 \) again using the chain rule. Let\[ v = (2 - 5x)^3. \] The derivative \( \frac{dv}{dx} \) using the chain rule is:1. Compute \( \frac{dv}{du} = 3u^2 \) where \( u = 2 - 5x. \)2. We know \( \frac{du}{dx} = -5 \) as from Step 2.\[ \frac{dv}{dx} = 3(2 - 5x)^2 \cdot (-5) = -15(2 - 5x)^2. \]Now, use product rule for \( \frac{d(-40v)}{dx} \):\[ \frac{d}{dx}(-40(2-5x)^3) = -40 \cdot (-15(2 - 5x)^2) = 600(2 - 5x)^2. \]
04
Find the Complete Expression
Combine factors to write the second derivative in conventional form:The second derivative is:\[ \frac{d^2y}{dx^2} = 600(2 - 5x)^2. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental technique used in calculus for finding derivatives of composite functions. If you have a function composed of an inner and an outer function, the chain rule allows you to differentiate by following a simple process.
In the original exercise, the function we need to differentiate is in the form \( y = 2(2 - 5x)^4 \). This is a composite function with the inner function being \( u = 2 - 5x \) and the outer function \( y = 2u^4 \). The chain rule helps us deal with these layers:
In the original exercise, the function we need to differentiate is in the form \( y = 2(2 - 5x)^4 \). This is a composite function with the inner function being \( u = 2 - 5x \) and the outer function \( y = 2u^4 \). The chain rule helps us deal with these layers:
- First, differentiate the outer function with respect to the inner function. This gives us \( \frac{dy}{du} = 8u^3 \).
- Next, differentiate the inner function with respect to the variable \( x \), giving \( \frac{du}{dx} = -5 \).
- Finally, multiply these derivatives together: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 8u^3 \cdot -5 \).
Product Rule
The product rule is another essential technique in calculus used when differentiating products of two functions. It simplifies the process by providing a method to take derivatives where multiplication is involved.
In our solution process, although mainly dominated by the chain rule, the product rule was subtly applied in the expression \( -40(2 - 5x)^3 \), when we moved to find the second derivative. Once we needed to differentiate \( -40 \cdot v \) (where \( v = (2 - 5x)^3 \)), the product rule became necessary:
In our solution process, although mainly dominated by the chain rule, the product rule was subtly applied in the expression \( -40(2 - 5x)^3 \), when we moved to find the second derivative. Once we needed to differentiate \( -40 \cdot v \) (where \( v = (2 - 5x)^3 \)), the product rule became necessary:
- If you have \( f(x) = u(x)\cdot v(x) \), then the derivative is \( f'(x) = u'(x)\cdot v(x) + u(x)\cdot v'(x) \).
- Applying this, for \( u = -40 \) (constant), and \( v = (2 - 5x)^3 \), the second term simplifies our process: \( -40 \cdot \frac{dv}{dx} \).
Binomial Function
The binomial function appears frequently in calculus problems involving powers of expressions. A binomial is simply an expression composed of two terms, typically written as \( (a + b)^n \) or \( (a - b)^n \). It serves as a backbone for many algebraic manipulations.
For our initial function, \( y = 2(2-5x)^4 \), the core lies in the raised binomial structure:
For our initial function, \( y = 2(2-5x)^4 \), the core lies in the raised binomial structure:
- Each context involving differentiation often begins with addressing the powers and coefficients within these binomials.
- By consistently applying the power rule in tandem with strategies like the chain rule, the differentiation of binomial expressions becomes effective.
- This function's power of 4 necessitates repetitive application in finding the derivatives. This ensures accuracy and consistency in higher-order differentiation.
