Problem 71
Question
Use the given information to find \(f^{\prime}(2)\) \(g(2)=3\) and \(g^{\prime}(2)=-2\) \(h(2)=-1 \quad\) and \(\quad h^{\prime}(2)=4\) $$ f(x)=g(x)+h(x) $$
Step-by-Step Solution
Verified Answer
The value of \(f'(2)\) is 2.
1Step 1: Understand the problem
The function \(f(x)\) is the sum of \(g(x)\) and \(h(x)\), so \(f'(x)\) will be the sum of \(g'(x)\) and \(h'(x)\). This is derived from the sum rule in differentiation.
2Step 2: Apply the sum rule
According to the sum rule, \(f'(x) = g'(x) + h'(x)\). We are given the values of \(g'(x)\) and \(h'(x)\) at x=2. So, \(f'(2) = g'(2) + h'(2)\).
3Step 3: substitute the given values
Substitute the given values into the equation from step 2. \(f'(2) = g'(2) + h'(2) = -2 + 4 = 2.\)
Key Concepts
Sum RuleDerivative of a FunctionEvaluation of Derivatives
Sum Rule
The Sum Rule in differentiation is a fundamental concept that makes calculating derivatives of functions much simpler. It states that if you have two differentiable functions, say \( g(x) \) and \( h(x) \), their sum \( f(x) = g(x) + h(x) \) is also differentiable. For the derivative of this sum, the Sum Rule tells us that:
To apply this, you do not need to reinvent the wheel for every problem. Instead, check that both functions are differentiable, and then add their derivatives to get the derivative of their sum.In our original exercise, we have \( f(x) = g(x) + h(x) \), where \( f'(x) \) was found by adding \( g'(x) \) and \( h'(x) \). This shows how efficiently the Sum Rule can be applied to find the derivative of combined functions.
- \( f'(x) = g'(x) + h'(x) \)
To apply this, you do not need to reinvent the wheel for every problem. Instead, check that both functions are differentiable, and then add their derivatives to get the derivative of their sum.In our original exercise, we have \( f(x) = g(x) + h(x) \), where \( f'(x) \) was found by adding \( g'(x) \) and \( h'(x) \). This shows how efficiently the Sum Rule can be applied to find the derivative of combined functions.
Derivative of a Function
The derivative of a function is essentially a measure of how a function's output changes as its input changes. More formally, if you have a function \( f(x) \), its derivative, denoted as \( f'(x) \), represents the rate of change or slope of \( f \) at any given point.
Here's a simple way to think about derivatives:
Here's a simple way to think about derivatives:
- They give you the instantaneous rate of change at any point on the function.
- They're used to find tangent lines, which tell the slope at a specific point.
- They help in understanding the behavior of the function, such as increasing or decreasing trends.
Evaluation of Derivatives
Evaluation of derivatives is the process of finding the derivative of a function at a particular point. In our exercise, we needed to evaluate the derivative \( f'(x) \) at \( x = 2 \), using the information given for \( g'(x) \) and \( h'(x) \) at this point.
Here's how you can evaluate derivatives effectively:
Here's how you can evaluate derivatives effectively:
- Substitute the given values for derivatives into the derivative expression.
- For our function \( f'(2) = g'(2) + h'(2) \), substitute \( g'(2) = -2 \) and \( h'(2) = 4 \).
- Simply calculate the sum of these given values:
Other exercises in this chapter
Problem 70
Find the derivative of the given function \(f\). Then use a graphing utility to graph \(f\) and its derivative in the same viewing window. What does the \(x\) -
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You deposit in an account that is compounded quarterly at an annual rate of \(r\) (in decimal form). The balance \(A\) after 10 years is \(A=2000\left(1+\frac{r
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The slope of the graph of \(y=x^{2}\) is different at every point on the graph of \(f\).
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Consider a certificate of deposit that pays \(10 \%\) (annual percentage rate) on an initial deposit of \(\$ 1000\). The balance \(A\) after 10 years is \(A=100
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