Problem 38
Question
Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=1\), assuming that \(f(1)=2\) and \(f^{\prime}(1)=-1\) \(y=3 x^{2} f(x)\)
Step-by-Step Solution
Verified Answer
The derivative of \(y\) at \(x=1\) is 9.
1Step 1: Identify the Function Structure
Recognize that the given function is a product of two functions: a polynomial function \(g(x) = 3x^2\) and a differentiable function \(h(x) = f(x)\). Hence, the function is \(y = g(x)h(x) = 3x^2 f(x)\).
2Step 2: Apply the Product Rule
The product rule states that if \(y = u(x)v(x)\), then \(y' = u'(x) v(x) + u(x) v'(x)\). Here, set \(u(x) = 3x^2\) and \(v(x) = f(x)\). Thus, the derivative \(y' = (3x^2)' f(x) + 3x^2 f'(x)\).
3Step 3: Differentiate the Components
Calculate the derivatives: the derivative of \(3x^2\) is \(6x\), and the derivative of \(f(x)\) is given as \(f'(x)\). Substitute these into the product rule result to get: \[ y' = 6x f(x) + 3x^2 f'(x) \]
4Step 4: Substitute Known Values
Substitute the values \(f(1) = 2\) and \(f'(1) = -1\) into the expression found in Step 3. Also substitute \(x = 1\) since the derivative is evaluated at \(x = 1\). \[ y' = 6(1)(2) + 3(1)^2(-1) \] \[ = 12 - 3 \]
5Step 5: Calculate the Derivative at x=1
Simplify the expression to determine \(y'\) evaluated at \(x=1\). The calculation from Step 4 simplifies to \[ y' = 12 - 3 = 9 \].
Key Concepts
The Product RuleUnderstanding Differentiable FunctionsPolynomial FunctionsFunction Substitution in Calculus
The Product Rule
In calculus, when dealing with the derivative of a product of two functions, the product rule is essential. If you have a function that is the product of two other functions, like our case with \( y = g(x)h(x) = 3x^2 f(x) \), you use the product rule to find the derivative.
Here's how the product rule works: If you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:
In our original problem, \( u(x) = 3x^2 \) and \( v(x) = f(x) \). Apply the product rule here and differentiate \( 3x^2 \) and then \( f(x) \). It ensures each part is accounted for when finding the total derivative.
Here's how the product rule works: If you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:
- \( y' = u'(x)v(x) + u(x)v'(x) \)
In our original problem, \( u(x) = 3x^2 \) and \( v(x) = f(x) \). Apply the product rule here and differentiate \( 3x^2 \) and then \( f(x) \). It ensures each part is accounted for when finding the total derivative.
Understanding Differentiable Functions
A differentiable function is simply a function that has a derivative at each point in its domain. This implies that at every point, the function has a well-defined tangent. For our exercise, knowing that \( f(x) \) is differentiable means that both \( f(x) \) and its derivative \( f'(x) \) exist.
Differentiability ensures that we can apply rules of calculus, such as the product rule, to find derivatives of more complex functions. When \( f(1) = 2 \) and \( f'(1) = -1 \), it indicates continuity and smooth behavior of the function around \( x = 1 \). Without this property, the derivative calculations might not be feasible, highlighting the importance of this concept.
Differentiability ensures that we can apply rules of calculus, such as the product rule, to find derivatives of more complex functions. When \( f(1) = 2 \) and \( f'(1) = -1 \), it indicates continuity and smooth behavior of the function around \( x = 1 \). Without this property, the derivative calculations might not be feasible, highlighting the importance of this concept.
Polynomial Functions
A polynomial function is a sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient. Our exercise involves a polynomial function of \( g(x) = 3x^2 \). Recognizing this structure helps to easily compute derivatives.
Thanks to its simplicity and predictable pattern, differentiating a polynomial like \( 3x^2 \) is straightforward — just bring down the exponent as a coefficient (\( 2 \times 3 \)) and reduce the power by one. This basic rule simplifies the product rule application, forming the backbone of derivative calculation.
Thanks to its simplicity and predictable pattern, differentiating a polynomial like \( 3x^2 \) is straightforward — just bring down the exponent as a coefficient (\( 2 \times 3 \)) and reduce the power by one. This basic rule simplifies the product rule application, forming the backbone of derivative calculation.
Function Substitution in Calculus
Function substitution involves replacing a variable within an expression with another function. This is crucial when applying the product rule, especially when knowing the function's values at certain points.
In our problem, after computing the derivative, we substitute \( x = 1 \), \( f(1) = 2 \), and \( f'(1) = -1 \) into \( y' = 6x f(x) + 3x^2 f'(x) \) to find the specific value of the derivative at \( x=1 \).
In our problem, after computing the derivative, we substitute \( x = 1 \), \( f(1) = 2 \), and \( f'(1) = -1 \) into \( y' = 6x f(x) + 3x^2 f'(x) \) to find the specific value of the derivative at \( x=1 \).
- \( y' = 6(1)(2) + 3(1)^2(-1) \)
- \( = 12 - 3 \)
- \( = 9 \)
Other exercises in this chapter
Problem 38
Find the derivative with respect to the independent variable. $$ f(x)=\tan x \cot x $$
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Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \frac{2 x}{1+x^{2}} $$
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Differentiate $$ g(N)=r N\left(1-\frac{N}{K}\right) $$ with respect to \(N\). Assume that \(K\) and \(r\) are positive constants.
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In Problems , graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ f(x)=
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