Problem 25
Question
Differentiate $$ f(x)=a x^{2}-2 a $$ with respect to \(x\). Assume that \(a\) is a constant.
Step-by-Step Solution
Verified Answer
The derivative of \( f(x) = a x^{2} - 2a \) is \( 2ax \).
1Step 1: Identify the Function
The function given is \( f(x) = a x^{2} - 2a \). We need to differentiate this function with respect to \( x \). Remember that \( a \) is a constant.
2Step 2: Differentiate Each Term
Differentiate the first term \( a x^{2} \). Use the power rule which states that \( \frac{d}{dx}(x^n) = nx^{n-1} \). Thus, the derivative of \( a x^{2} \) is \( 2ax \).
3Step 3: Differentiate the Constant Term
The second term \(-2a\) is a constant with respect to \( x \). The derivative of any constant is 0, so the derivative of \(-2a\) is 0.
4Step 4: Combine Derivatives
Combine the results from the previous steps. The derivative of \( f(x) = a x^{2} - 2a \) with respect to \( x \) is \( 2ax + 0 \), which simplifies to \( 2ax \).
Key Concepts
Understanding the Power Rule in DifferentiationThe Role of Constant Terms in DifferentiationDerivative Computation Simplified
Understanding the Power Rule in Differentiation
The power rule is a fundamental technique for finding derivatives. It comes in handy when differentiating terms that involve powers of a variable. In essence, the power rule tells us how to take derivatives of expressions like \(x^n\). It states:
For example, in the term \(a x^2\), the exponent \(n\) equals 2. Applying the power rule, \(2 \times ax^{2-1} = 2ax\). Here, the variable \(a\) remains because it acts as a coefficient—it's just there to scale the function.
- \( \frac{d}{dx}(x^n) = nx^{n-1} \)
For example, in the term \(a x^2\), the exponent \(n\) equals 2. Applying the power rule, \(2 \times ax^{2-1} = 2ax\). Here, the variable \(a\) remains because it acts as a coefficient—it's just there to scale the function.
The Role of Constant Terms in Differentiation
Constant terms are simply numbers that don't change no matter what the variable is. In differentiation, dealing with constants is straightforward. For instance, consider the term \(-2a\) in the function. Since \(a\) is a constant, \(-2a\) behaves as one.
When you differentiate a constant, the result is always zero no matter what the constant is. This is because constants have no rate of change with respect to the variable we are differentiating. Hence, in this function, the derivative of the constant term \(-2a\) becomes 0.
So, no matter how complex the rest of your function gets, you can always rely on this rule to simplify the process at any stage of differentiation.
When you differentiate a constant, the result is always zero no matter what the constant is. This is because constants have no rate of change with respect to the variable we are differentiating. Hence, in this function, the derivative of the constant term \(-2a\) becomes 0.
So, no matter how complex the rest of your function gets, you can always rely on this rule to simplify the process at any stage of differentiation.
Derivative Computation Simplified
Once you understand how to differentiate terms using the power rule and recognize constant terms, putting it all together to compute a derivative becomes a task of follow-through.
Here’s a recap with our function \(f(x) = ax^2 - 2a\):
This simplifies to the neatly compact form of \(2ax\). By taking each part of a function, applying the right rule, and combining the results, you can simplify even the most complex-looking problems into manageable steps.
Here’s a recap with our function \(f(x) = ax^2 - 2a\):
- First, apply the power rule to \(ax^2\), resulting in \(2ax\).
- Next, recognize \(-2a\) as a constant term, resulting in a derivative of 0.
- Combine these results to find the overall derivative, \(2ax + 0\).
This simplifies to the neatly compact form of \(2ax\). By taking each part of a function, applying the right rule, and combining the results, you can simplify even the most complex-looking problems into manageable steps.
Other exercises in this chapter
Problem 25
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln (1-2 x) $$
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Apply the product rule for the product of three functions to find the derivative of \(y=f(x)\). \(f(x)=(x-3)(2-3 x)(5-x)\)
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If \(f(x)\) is differentiable for all \(x \in \mathbf{R}\) except at \(x=c\), is it true that \(f(x)\) must be continuous at \(x=c\) ? Justify your answer.
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Two people start biking from the same point. One heads east at \(15 \mathrm{mph}\), the other south at \(18 \mathrm{mph}\). What is the rate at which the distan
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