Calculus: Find Dy/dx For Cos(x^3 Y^3) = 5y^3

Hey calculus adventurers! Today, we're diving deep into the fascinating world of implicit differentiation to solve a gnarly problem: finding $\frac{dy}{dx}$ when we're given the equation $\cos \left(x^3 y^3\right)=5 y^3$. This might look a bit intimidating at first glance, but trust me, guys, with a bit of strategic thinking and a solid understanding of the chain rule, we'll break it down piece by piece. Implicit differentiation is our superpower here because we have y intertwined with x in a way that makes it super difficult, if not impossible, to isolate y on one side of the equation. Instead of fighting to get y by itself, we're going to treat y as a function of x and differentiate both sides of the equation with respect to x. Remember, whenever we differentiate a term involving y, we've got to multiply by $\frac{dy}{dx}$ – that's the chain rule in action, and it's absolutely crucial for these types of problems. So, grab your pencils, maybe a coffee, and let's get ready to unravel this mathematical mystery together. We'll be using the product rule and the chain rule extensively, so if those are a bit rusty, now's a great time for a quick refresher. The goal is to isolate $\frac{dy}{dx}$ in the end, so we'll be doing a lot of algebraic manipulation, moving terms around, and factoring. It's like a puzzle, and each step brings us closer to the solution. Don't get discouraged if it looks complex; breaking it down into smaller, manageable steps is the key to conquering these calculus challenges. We're going to start by differentiating the left side of the equation, then the right side, and then bring it all together to solve for our precious $\frac{dy}{dx}$. Ready to get your calculus on?

Step 1: Differentiating Both Sides with Respect to x

Alright, let's get down to business. We have our equation: $\cos \left(x^3 y^3\right)=5 y^3$. Our first major move is to differentiate both sides of this equation with respect to x. This is where the magic of implicit differentiation truly begins. On the left side, $\cos \left(x^3 y^3\right)$, we need to use the chain rule because we have a function within a function. The outer function is cos(u), and the inner function is u = x^3 y^3. The derivative of cos(u) with respect to u is -sin(u). Now, we need to find the derivative of the inner function, $x^3 y^3$, with respect to x. This requires the product rule, because we have x^3 multiplied by y^3. The product rule states that if we have f(x)g(x), its derivative is f'(x)g(x) + f(x)g'(x). Here, let $f(x) = x^3$ and $g(x) = y^3$. The derivative of $f(x)$ is $f'(x) = 3x^2$. For $g(x) = y^3$, its derivative with respect to x is $g'(x) = 3y^2 \frac{dy}{dx}$ (remember that crucial chain rule factor for y!). So, applying the product rule to $x^3 y^3$, we get: $(3x^2)(y^3) + (x^3)(3y^2 \frac{dy}{dx}) = 3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}$.

Now, putting it all together for the left side of our original equation, the derivative of $\cos \left(x^3 y^3\right)$ with respect to x is: $-\sin \left(x^3 y^3\right) \cdot \left(3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}\right)$.

Moving over to the right side of the equation, we have $5y^3$. Again, we're differentiating with respect to x. This is a simpler application of the chain rule. The derivative of $5y^3$ with respect to y is $15y^2$. Since y is a function of x, we multiply by $\frac{dy}{dx}$. So, the derivative of $5y^3$ with respect to x is $15y^2 \frac{dy}{dx}$.

Now we equate the derivatives of both sides: $-\sin \left(x^3 y^3\right) \cdot \left(3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}\right) = 15y^2 \frac{dy}{dx}$. This is our pivotal equation from which we will isolate $\frac{dy}{dx}$. Keep your eyes on the prize, guys!

Step 2: Expanding and Rearranging the Equation

Okay, calculus crusaders, we've got our differentiated equation: $-\sin \left(x^3 y^3\right) \cdot \left(3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}\right) = 15y^2 \frac{dy}{dx}$. Our next mission, should we choose to accept it, is to simplify this beast and start isolating that $\frac{dy}{dx}$. First, let's distribute that $-\sin \left(x^3 y^3\right)$ term on the left side. This will help us separate the terms that contain $\frac{dy}{dx}$ from those that don't. So, we multiply $-\sin \left(x^3 y^3\right)$ by $3x^2 y^3$, and then by $3x^3 y^2 \frac{dy}{dx}$.

This gives us: $-3x^2 y^3 \sin \left(x^3 y^3\right) - 3x^3 y^2 \sin \left(x^3 y^3\right) \frac{dy}{dx} = 15y^2 \frac{dy}{dx}$.

See how we now have $\frac{dy}{dx}$ terms on both sides of the equation? Our goal is to gather all the terms containing $\frac{dy}{dx}$ on one side and all the other terms on the other side. It's a classic algebra move, but super important here. Let's move the term $-3x^3 y^2 \sin \left(x^3 y^3\right) \frac{dy}{dx}$ from the left side to the right side. When we move it across the equals sign, its sign flips.

So, the equation becomes: $-3x^2 y^3 \sin \left(x^3 y^3\right) = 15y^2 \frac{dy}{dx} + 3x^3 y^2 \sin \left(x^3 y^3\right) \frac{dy}{dx}$.

Now, look at the right side. Both terms have $\frac{dy}{dx}$. This is a prime opportunity to use factoring! We can pull $\frac{dy}{dx}$ out as a common factor. This is a game-changer for isolating our derivative.

Factoring out $\frac{dy}{dx}$ from the right side, we get: $-3x^2 y^3 \sin \left(x^3 y^3\right) = \frac{dy}{dx} \left(15y^2 + 3x^3 y^2 \sin \left(x^3 y^3\right)\right)$.

We're so close, guys! The equation is now structured perfectly for us to perform the final step: dividing to solve for $\frac{dy}{dx}$. Keep that positive energy going; we're navigating this complex calculus terrain like pros!

Step 3: Isolating dy/dx

We've reached the home stretch, my fellow mathematicians! Our equation is: $-3x^2 y^3 \sin \left(x^3 y^3\right) = \frac{dy}{dx} \left(15y^2 + 3x^3 y^2 \sin \left(x^3 y^3\right)\right)$. The final step to find $\frac{dy}{dx}$ is pure algebraic wizardry. We simply need to isolate $\frac{dy}{dx}$. To do this, we'll divide both sides of the equation by the entire expression in the parentheses on the right side: $\left(15y^2 + 3x^3 y^2 \sin \left(x^3 y^3\right)\right)$.

Performing this division, we get:

$$\frac{dy}{dx} = \frac{-3x^2 y^3 \sin \left(x^3 y^3\right)}{15y^2 + 3x^3 y^2 \sin \left(x^3 y^3\right)}$$

And there you have it! We have successfully found $\frac{dy}{dx}$ in terms of $x$ and $y$. However, we can often simplify expressions like this further. Let's take a look at the numerator and the denominator. Notice that both terms in the denominator have a common factor of $3y^2$. We can also factor $3y^2$ out of the numerator. Let's factor it out from both the numerator and the denominator to see if we can cancel anything out.

Factor out $3y^2$ from the numerator: $-3x^2 y^3 \sin \left(x^3 y^3\right) = 3y^2 \left(-x^2 y \sin \left(x^3 y^3\right)\right)$.

Factor out $3y^2$ from the denominator: $15y^2 + 3x^3 y^2 \sin \left(x^3 y^3\right) = 3y^2 \left(5 + x^3 \sin \left(x^3 y^3\right)\right)$.

Now, substitute these back into our expression for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{3y^2 \left(-x^2 y \sin \left(x^3 y^3\right)\right)}{3y^2 \left(5 + x^3 \sin \left(x^3 y^3\right)\right)}$$

We can now cancel out the common factor of $3y^2$ (assuming $y \neq 0$). This gives us our final, simplified answer:

$$\boxed{\frac{dy}{dx} = \frac{-x^2 y \sin \left(x^3 y^3\right)}{5 + x^3 \sin \left(x^3 y^3\right)}}$$

Isn't that neat, guys? We took a seemingly complex equation and, by systematically applying the rules of differentiation and algebra, arrived at a clean expression for the derivative. This process highlights the power and elegance of calculus in describing the relationships between variables, even when they're tangled up together. Keep practicing, and you'll master these techniques in no time! Your brain will thank you for the workout!