Calculus Derivatives And Second Derivatives Problems And Solutions

This article explores several problems related to derivatives and second derivatives in calculus. We'll walk through the solutions step-by-step, providing clear explanations and insights along the way. Whether you're a student learning calculus or someone looking to refresh your knowledge, this guide will help you understand key concepts and techniques. Let's dive into these calculus problems and see how to tackle them effectively.

1. Finding the Derivative of $y = \cos^2 x + \sin^2 x$

Understanding Trigonometric Identities and Differentiation

To find $\frac{dy}{dx}$ for $y = \cos^2 x + \sin^2 x$, we can use a fundamental trigonometric identity to simplify the expression before differentiating. This approach not only makes the differentiation process easier but also provides a great example of how leveraging identities can streamline calculus problems. Our main keyword here is derivative of trigonometric functions. The identity we'll use is the Pythagorean identity, which states that $\cos^2 x + \sin^2 x = 1$. This identity is a cornerstone of trigonometry and is essential for simplifying expressions involving sine and cosine functions. So, we start with our given function:

$y = \cos^2 x + \sin^2 x$

Applying the Pythagorean identity, we get:

$y = 1$

Now that we have simplified the function, differentiating it becomes straightforward. The derivative of a constant is always zero. This is a basic rule in calculus and applies to any constant value, including 1. Therefore, we have:

$\frac{dy}{dx} = \frac{d}{dx}(1) = 0$

This result shows that the rate of change of $y$ with respect to $x$ is zero, which makes sense since $y$ is a constant function. In summary, to find the derivative of $y = \cos^2 x + \sin^2 x$, we first used the trigonometric identity to simplify the function to $y = 1$, and then we differentiated the constant function to find $\frac{dy}{dx} = 0$. This problem underscores the importance of simplifying expressions before differentiation, which can often save time and reduce the complexity of the calculation. Remember, recognizing and applying appropriate trigonometric identities is a powerful tool in calculus.

2. Finding $\frac{dy}{dx}$ for $3 + y^2 = x^3 - y$

Implicit Differentiation Explained

In this problem, we need to find $\frac{dy}{dx}$ for the equation $3 + y^2 = x^3 - y$. This equation is an example of an implicit function, where $y$ is not explicitly defined in terms of $x$. To solve this, we'll use a technique called implicit differentiation. Implicit differentiation involves differentiating both sides of the equation with respect to $x$, treating $y$ as a function of $x$. This means we'll need to apply the chain rule when differentiating terms involving $y$. Our main keyword here is implicit differentiation. Let's start by differentiating both sides of the equation with respect to $x$:

$\frac{d}{dx}(3 + y^2) = \frac{d}{dx}(x^3 - y)$

Now, we differentiate each term separately. The derivative of a constant (3) is 0. For $y^2$, we apply the chain rule, which gives us $2y \frac{dy}{dx}$. On the right side, the derivative of $x^3$ is $3x^2$, and the derivative of $-y$ is $-\frac{dy}{dx}$. So, we have:

$0 + 2y \frac{dy}{dx} = 3x^2 - \frac{dy}{dx}$

Next, we need to isolate $\frac{dy}{dx}$. To do this, we'll move all terms involving $\frac{dy}{dx}$ to one side of the equation and the remaining terms to the other side:

$2y \frac{dy}{dx} + \frac{dy}{dx} = 3x^2$

Now, we can factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx}(2y + 1) = 3x^2$

Finally, we divide both sides by $(2y + 1)$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{3x^2}{2y + 1}$

This is the derivative of $y$ with respect to $x$ for the given implicit function. In summary, we used implicit differentiation to find $\frac{dy}{dx}$. This involved differentiating both sides of the equation with respect to $x$, applying the chain rule where necessary, and then isolating $\frac{dy}{dx}$. Implicit differentiation is a powerful tool for finding derivatives of functions that are not explicitly defined. It is crucial to remember the chain rule when differentiating terms involving $y$.

3. Finding $y''(0)$ for $y(x) = 3e^{2x} + \sin x$

Calculating Second Derivatives

To find $y''(0)$ for the function $y(x) = 3e^{2x} + \sin x$, we need to calculate the second derivative of $y$ with respect to $x$ and then evaluate it at $x = 0$. The second derivative is the derivative of the first derivative, so we'll need to differentiate the function twice. Our main keyword here is second derivative. Let's start by finding the first derivative, $y'(x)$. We'll differentiate each term separately. For $3e^{2x}$, we use the chain rule, which states that the derivative of $e^{f(x)}$ is $f'(x)e^{f(x)}$. In this case, $f(x) = 2x$, so $f'(x) = 2$. Thus, the derivative of $3e^{2x}$ is $3 \cdot 2e^{2x} = 6e^{2x}$. The derivative of $\sin x$ is $\cos x$. Therefore, the first derivative is:

$y'(x) = 6e^{2x} + \cos x$

Now, we need to find the second derivative, $y''(x)$. We differentiate $y'(x)$ with respect to $x$. For $6e^{2x}$, we again use the chain rule, which gives us $6 \cdot 2e^{2x} = 12e^{2x}$. The derivative of $\cos x$ is $-\sin x$. So, the second derivative is:

$y''(x) = 12e^{2x} - \sin x$

Finally, we evaluate $y''(x)$ at $x = 0$:

$y''(0) = 12e^{2(0)} - \sin(0)$

Since $e^{2(0)} = e^0 = 1$ and $\sin(0) = 0$, we have:

$y''(0) = 12(1) - 0 = 12$

Therefore, the value of the second derivative at $x = 0$ is 12. In summary, to find $y''(0)$, we first calculated the first derivative $y'(x)$, then we found the second derivative $y''(x)$, and finally, we evaluated $y''(x)$ at $x = 0$. This problem highlights the process of finding higher-order derivatives and evaluating them at specific points. Remember to apply the chain rule correctly when differentiating exponential functions and to know the basic derivatives of trigonometric functions.

4. Finding $\frac{dy}{dx}$ for $y = 6xe^{-5x}$

Applying the Product Rule and Chain Rule

To find $\frac{dy}{dx}$ for $y = 6xe^{-5x}$, we need to use the product rule since we have a product of two functions: $6x$ and $e^{-5x}$. The product rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx} = u'v + uv'$. Additionally, we'll need to use the chain rule to differentiate $e^{-5x}$. Our main keyword here is product rule. Let's identify $u$ and $v$:

$u = 6x$ $v = e^{-5x}$

Now, we find the derivatives of $u$ and $v$ with respect to $x$:

$u' = \frac{d}{dx}(6x) = 6$

To find $v'$, we use the chain rule. The derivative of $e^{f(x)}$ is $f'(x)e^{f(x)}$. In this case, $f(x) = -5x$, so $f'(x) = -5$. Thus, the derivative of $e^{-5x}$ is $-5e^{-5x}$:

$v' = \frac{d}{dx}(e^{-5x}) = -5e^{-5x}$

Now, we apply the product rule:

$\frac{dy}{dx} = u'v + uv' = 6e^{-5x} + 6x(-5e^{-5x})$

Simplify the expression:

$\frac{dy}{dx} = 6e^{-5x} - 30xe^{-5x}$

We can factor out $6e^{-5x}$ to further simplify:

$\frac{dy}{dx} = 6e^{-5x}(1 - 5x)$

This is the derivative of $y$ with respect to $x$. In summary, to find $\frac{dy}{dx}$ for $y = 6xe^{-5x}$, we applied the product rule and the chain rule. The product rule was necessary because we had a product of two functions, and the chain rule was needed to differentiate the exponential function. Remember to identify the functions $u$ and $v$ correctly and to apply the rules of differentiation carefully. Simplifying the expression after applying the rules is also an important step.

5. Finding the Derivative of...

General Strategies for Differentiation

This question is incomplete, but let's discuss general strategies for finding derivatives. The derivative of a function represents its instantaneous rate of change, and mastering differentiation is crucial in calculus. Our main keyword here is differentiation techniques. Depending on the function, different rules and techniques may be required. Here are some common strategies:

1. Basic Differentiation Rules:

   • Power Rule: If $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$. This rule is fundamental and applies to polynomial functions.
   • Constant Multiple Rule: If $y = cf(x)$, where $c$ is a constant, then $\frac{dy}{dx} = cf'(x)$. This rule allows us to handle constant coefficients.
   • Sum/Difference Rule: If $y = u(x) \pm v(x)$, then $\frac{dy}{dx} = u'(x) \pm v'(x)$. This rule allows us to differentiate term by term.
   • Constant Rule: If $y = c$, where $c$ is a constant, then $\frac{dy}{dx} = 0$.

2. Product Rule:

   • As we discussed in problem 4, if $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$. The product rule is essential for differentiating functions that are products of other functions.

3. Quotient Rule:

   • If $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. The quotient rule is used for differentiating functions that are quotients of other functions.

4. Chain Rule:

   • If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x))g'(x)$. The chain rule is crucial for differentiating composite functions, where one function is nested inside another.

5. Derivatives of Trigonometric Functions:

   • $\frac{d}{dx}(\sin x) = \cos x$
   • $\frac{d}{dx}(\cos x) = -\sin x$
   • $\frac{d}{dx}(\tan x) = \sec^2 x$
   • $\frac{d}{dx}(\csc x) = -\csc x \cot x$
   • $\frac{d}{dx}(\sec x) = \sec x \tan x$
   • $\frac{d}{dx}(\cot x) = -\csc^2 x$

6. Derivatives of Exponential and Logarithmic Functions:

   • $\frac{d}{dx}(e^x) = e^x$
   • $\frac{d}{dx}(a^x) = a^x \ln a$
   • $\frac{d}{dx}(\ln x) = \frac{1}{x}$
   • $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

7. Implicit Differentiation:

   • As we saw in problem 2, implicit differentiation is used when $y$ is not explicitly defined in terms of $x$. Differentiate both sides of the equation with respect to $x$ and use the chain rule for terms involving $y$.

8. Higher-Order Derivatives:

   • The second derivative, $y''$, is the derivative of the first derivative, $y'$. Similarly, higher-order derivatives can be found by differentiating successively.

In summary, finding the derivative of a function requires understanding and applying the appropriate differentiation rules and techniques. It is important to identify the structure of the function and choose the most efficient method. Practice is key to mastering differentiation. By understanding these strategies, you can approach a wide range of differentiation problems effectively.

By exploring these problems, we've covered various aspects of differentiation, from basic rules to more advanced techniques like implicit differentiation and the chain rule. Remember, practice is essential for mastering these concepts. Keep working on different types of problems, and you'll build a strong foundation in calculus.