Problem 29
Question
Find the derivative with respect to the independent variable. $$ f(x)=\sqrt{\sin \left(2 x^{2}-1\right)} $$
Step-by-Step Solution
Verified Answer
\( \frac{2x \cos(2x^2 - 1)}{\sqrt{\sin(2x^2 - 1)}} \)
1Step 1: Identify the Function
We are given the function \( f(x) = \sqrt{\sin \left(2x^2 - 1\right)} \), which is composed of an outer square root function and an inner sine function.
2Step 2: Apply Chain Rule - Outer Function
To differentiate the outer function, \( u(x) = \sqrt{u} = u^{\frac{1}{2}} \), apply the power rule: \( \frac{d}{du} u^{1/2} = \frac{1}{2}u^{-1/2} \). Let \( u = \sin(2x^2 - 1) \). So, \( \frac{d}{dx} f(x) = \frac{1}{2} \sin(2x^2 - 1)^{-1/2} \cdot \frac{d}{dx}(\sin(2x^2 - 1)) \).
3Step 3: Apply Chain Rule - Inner Function
Differentiate the inner function, \( \sin(v) \), where \( v = 2x^2 - 1 \). The derivative is \( \cos(v) \cdot \frac{d}{dx}(2x^2 - 1) \).
4Step 4: Differentiate Known Expression
The derivative of \( 2x^2 - 1 \) with respect to \( x \) is \( 4x \). Thus, the derivative of the inner function \( \sin(2x^2 - 1) \) is \( \cos(2x^2 - 1) \times 4x \).
5Step 5: Combine Results
Substitute the derivative of the inner function back into the expression found in Step 2: \[ \frac{d}{dx} f(x) = \frac{1}{2} \sin(2x^2 - 1)^{-1/2} \cdot (\cos(2x^2 - 1) \cdot 4x) \].
6Step 6: Simplify the Expression
Simplify the expression: \( \frac{d}{dx} f(x) = \frac{4x \cos(2x^2 - 1)}{2\sqrt{\sin(2x^2 - 1)}} \). This further simplifies to \( \frac{2x \cos(2x^2 - 1)}{\sqrt{\sin(2x^2 - 1)}} \).
Key Concepts
Chain RulePower RuleTrigonometric Derivatives
Chain Rule
The Chain Rule is a method for differentiating compositions of functions. It is a vital technique in calculus, especially when dealing with complex functions. Think of it like peeling an onion, where you handle one layer (or function) at a time.
The Chain Rule states that if you have a function composed of two functions, say \( h(x) = f(g(x)) \), then the derivative of \( h \) is the derivative of the outer function \( f \) evaluated at \( g(x) \) times the derivative of \( g(x) \).
For example, in the function \( f(x) = \sqrt{\sin(2x^2 - 1)} \), there is an outer square root function and an inner sine function.
The Chain Rule states that if you have a function composed of two functions, say \( h(x) = f(g(x)) \), then the derivative of \( h \) is the derivative of the outer function \( f \) evaluated at \( g(x) \) times the derivative of \( g(x) \).
For example, in the function \( f(x) = \sqrt{\sin(2x^2 - 1)} \), there is an outer square root function and an inner sine function.
- First, we differentiate the outer function \( u = \sqrt{u} \) using the power rule for \( u^{1/2} \), resulting in \( \frac{1}{2}u^{-1/2} \).
- Next, we multiply by the derivative of the inner function, which involves differentiating \( \sin(v) \) using known trigonometric derivatives and applying the chain rule again for \( v = 2x^2 - 1 \).
Power Rule
The Power Rule provides a quick and straightforward way to differentiate functions in the form of \( x^n \). Specifically, if you have \( f(x) = x^n \), the derivative \( f'(x) \) is given by \( nx^{n-1} \).
In the context of our function, where \( u = \sin(2x^2 - 1) \) and we are considering \( u^{1/2} \), we apply the Power Rule to the expression \( u^{1/2} \). This results in \( \frac{1}{2}u^{-1/2} \), showing how the Power Rule makes it easy to handle such expressions.
Remember, the Power Rule is valuable for many derivatives:
In the context of our function, where \( u = \sin(2x^2 - 1) \) and we are considering \( u^{1/2} \), we apply the Power Rule to the expression \( u^{1/2} \). This results in \( \frac{1}{2}u^{-1/2} \), showing how the Power Rule makes it easy to handle such expressions.
Remember, the Power Rule is valuable for many derivatives:
- It's particularly helpful when you need to convert roots into powers since \( \sqrt{u} = u^{1/2} \) is a common conversion.
- Enables faster calculations compared to manually differentiating products or composite functions like those requiring the Chain Rule.
Trigonometric Derivatives
Trigonometric derivatives are particularly important when you have functions involving sine, cosine, or other trigonometric functions. For instance, knowing that the derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( \cos(x) \) is \(-\sin(x) \) is crucial.
In our example, we had to differentiate \( \sin(2x^2 - 1) \). Here’s how:
In our example, we had to differentiate \( \sin(2x^2 - 1) \). Here’s how:
- First, identify \( v = 2x^2 - 1 \), making \( \sin(v) \) the trigonometric part to address.
- The derivative of \( \sin(v) \) becomes \( \cos(v) \), a straightforward application of trigonometric derivatives.
- Next, apply the chain rule again for the function \( v = 2x^2 - 1 \) as we did, ultimately resulting in multiplying \( \cos(v) \) by \( 4x \), since the derivative of \( 2x^2 - 1 \) is \( 4x \).
Other exercises in this chapter
Problem 29
Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\sqrt{1+x^{2}}\) at \(a=0\)
View solution Problem 29
Find the equation of the tangent line to the curve \(y=\sqrt{x}\) at the point \((4,2)\).
View solution Problem 29
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left(2 x^{3}-x\right) $$
View solution Problem 29
Differentiate $$ f(x)=a(x+1)(2 x-1) $$ with respect to \(x\). Assume that \(a\) is a positive constant.
View solution