Mastering Derivatives: A Step-by-Step Guide

Hey guys! Derivatives can seem intimidating, but trust me, they're super useful in math and science. In this article, we're going to break down how to find derivatives, especially when things get a little more complex. We will walk through some examples step by step, focusing on making sure you understand the why behind the how. We'll tackle problems involving inverse trigonometric functions and implicit differentiation. Let's get started and make derivatives less scary and more manageable!

Part 1: Derivative of $f(x) = \sin^{-1}x + 3^{2x^2}$

Understanding the Problem

Okay, so we've got this function: $f(x) = \sin^{-1}x + 3^{2x^2}$. This looks a bit complex because it combines an inverse trigonometric function ($\sin^{-1}x$) with an exponential function ($3^{2x^2}$). To find the derivative, we'll need to use a couple of rules: the derivative of $\sin^{-1}x$ and the chain rule for the exponential part. Don't worry, we'll take it slow and make sure we get it.

Step-by-Step Solution

1. Break it Down: Think of $f(x)$ as two separate functions: $u(x) = \sin^{-1}x$ and $v(x) = 3^{2x^2}$. So, $f(x) = u(x) + v(x)$.

2. Derivative of $\bf{\sin^{-1}x}$: The derivative of $\sin^{-1}x$ is a standard result, which is $\frac{1}{\sqrt{1 - x^2}}$. This is something you might want to memorize or have handy in your notes.

3. Derivative of $\bf{3^{2x^2}}$: This is where the chain rule comes in. The chain rule is used when you have a function inside another function. Here, we have $2x^2$ inside the exponential function $3^x$. The chain rule says that if you have $y = f(g(x))$, then $y' = f'(g(x)) \\cdot g'(x)$.

   • Let $g(x) = 2x^2$. Then $v(x) = 3^{g(x)}$.
   • The derivative of $3^u$ with respect to $u$ is $3^u \\cdot \ln(3)$. So, the derivative of $3^{2x^2}$ with respect to $2x^2$ is $3^{2x^2} \\cdot \ln(3)$.
   • Now, we need the derivative of $g(x) = 2x^2$, which is $g'(x) = 4x$.
   • Using the chain rule, the derivative of $v(x) = 3^{2x^2}$ is $3^{2x^2} \\cdot \ln(3) \\cdot 4x$.

4. Combine the Derivatives: Now we add the derivatives of $u(x)$ and $v(x)$ to get the derivative of $f(x)$:

   $\bf{f'(x) = \frac{1}{\sqrt{1 - x^2}} + 4x \\cdot 3^{2x^2} \\cdot \ln(3)}$

Key Takeaways

• Remember the derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1 - x^2}}$.
• The chain rule is essential when dealing with composite functions (functions inside functions).
• For exponential functions like $a^{f(x)}$, the derivative involves the natural logarithm, $\ln(a)$.

Part 2: Implicit Differentiation of $3x^2 - 7y^2 + 4xy - 8x = 0$

Understanding Implicit Differentiation

Implicit differentiation is a technique we use when we have an equation where $y$ isn't explicitly defined as a function of $x$. In other words, we can't easily write $y = f(x)$. Instead, we have a relationship between $x$ and $y$, like our equation: $3x^2 - 7y^2 + 4xy - 8x = 0$. The trick here is to differentiate each term with respect to $x$, and whenever we differentiate a term involving $y$, we tack on a $\frac{dy}{dx}$ because of the chain rule.

Step-by-Step Solution

1. Differentiate Each Term: Let's go through the equation term by term:

   • The derivative of $3x^2$ with respect to $x$ is $6x$.

   • The derivative of $-7y^2$ with respect to $x$ is $-14y\\frac{dy}{dx}$. Notice the $\frac{dy}{dx}$ here because we're differentiating a $y$ term.

   • The derivative of $4xy$ with respect to $x$ needs the product rule. The product rule says that if you have $y = u(x)v(x)$, then $y' = u'(x)v(x) + u(x)v'(x)$.

     • Let $u(x) = 4x$ and $v(x) = y$.
     • Then $u'(x) = 4$ and $v'(x) = \frac{dy}{dx}$.
     • So, the derivative of $4xy$ is $4y + 4x\\frac{dy}{dx}$.

   • The derivative of $-8x$ with respect to $x$ is $-8$.

   • The derivative of 0 is 0.

2. Put it All Together: Combining these derivatives, we get:

   $\bf{6x - 14y\\frac{dy}{dx} + 4y + 4x\\frac{dy}{dx} - 8 = 0}$

3. Isolate $\bf{\frac{dy}{dx}}$: Now we need to get all the terms with $\frac{dy}{dx}$ on one side and everything else on the other side.

   • Rearrange the equation: $(-14y + 4x)\\frac{dy}{dx} = -6x - 4y + 8$
   • Divide to solve for $\frac{dy}{dx}$: $\bf{\frac{dy}{dx} = \frac{-6x - 4y + 8}{-14y + 4x}}$

4. Simplify (Optional): We can simplify this a bit by dividing everything by 2:

   $\bf{\frac{dy}{dx} = \frac{-3x - 2y + 4}{-7y + 2x}}$

Key Takeaways

• Implicit differentiation is used when $y$ isn't explicitly defined as a function of $x$.
• Remember to use the chain rule and add $\frac{dy}{dx}$ whenever you differentiate a term involving $y$.
• The product rule is often needed when you have terms like $xy$.
• Isolate $\frac{dy}{dx}$ to find the derivative.

Conclusion

So, there you have it! We've tackled finding derivatives of some pretty complex functions, including those with inverse trigonometric functions and implicit differentiation. The main thing to remember is to break the problem down into smaller, manageable steps, use the appropriate rules (like the chain rule and product rule), and take your time. Derivatives might seem tough at first, but with practice, you'll get the hang of it. Keep practicing, and you'll be a derivative pro in no time! Remember, understanding the why behind the how is key to mastering calculus concepts. Keep up the great work, guys!