Implicit Differentiation: Finding Tangent Line Slopes

by SLV Team 54 views

Hey everyone! Today, we're diving deep into the world of calculus, specifically, exploring implicit differentiation and how it helps us find the slope of a tangent line to a curve. We'll be working with the curve defined by the equation yx+4y=x7−5\frac{y}{x+4y} = x^7 - 5 and we want to find the slope of the tangent line at the point (1,−417)\left(1, \frac{-4}{17}\right). It might seem a little intimidating at first, but trust me, it's not as scary as it looks. Let's break it down step-by-step. Remember, the slope of a tangent line gives us the instantaneous rate of change of the function at a specific point. So, what exactly is implicit differentiation, and why do we need it? Let's get into it, guys!

Understanding Implicit Differentiation

So, what's the deal with implicit differentiation? Well, sometimes, the relationship between xx and yy isn't explicitly given in the form of y=f(x)y = f(x). Instead, we have an equation where xx and yy are mixed together. This is where implicit differentiation comes in handy. It's a technique that allows us to find the derivative of yy with respect to xx even when yy isn't isolated. Basically, when we differentiate implicitly, we treat yy as a function of xx. Whenever we differentiate a term involving yy, we have to remember to multiply by dydx\frac{dy}{dx}, which represents the derivative of yy with respect to xx. This is super important! The chain rule is the secret ingredient here. It's what allows us to handle those tricky yy terms. Think of it like this: when we differentiate y2y^2, we get 2y2y, but because yy is a function of xx, we also need to multiply by dydx\frac{dy}{dx}, resulting in 2ydydx2y\frac{dy}{dx}. That dydx\frac{dy}{dx} is the slope we're after, the slope of that tangent line! It's the key to unlocking the problem, so don't miss it! Now, let's get our hands dirty and actually solve the problem. Get ready to put on your thinking caps, because we're about to apply these concepts to our specific curve and point. Trust me, it's going to be worth it when we find that slope!

Differentiating the Equation

Alright, let's get down to business! The first step is to differentiate both sides of the equation yx+4y=x7−5\frac{y}{x+4y} = x^7 - 5 with respect to xx. This is where things get interesting, so stick with me! On the left side, we have a fraction, so we'll need to use the quotient rule, which states that the derivative of uv\frac{u}{v} is vdudx−udvdxv2\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. Let's break this down. In our case, u=yu = y and v=x+4yv = x + 4y. Therefore, dudx=dydx\frac{du}{dx} = \frac{dy}{dx} and dvdx=1+4dydx\frac{dv}{dx} = 1 + 4\frac{dy}{dx}. Applying the quotient rule, we get:

(x+4y)dydx−y(1+4dydx)(x+4y)2\frac{(x+4y)\frac{dy}{dx} - y(1 + 4\frac{dy}{dx})}{(x+4y)^2}.

On the right side of the equation, the derivative of x7−5x^7 - 5 is simply 7x67x^6, since the derivative of a constant is zero. Therefore, our differentiated equation is:

(x+4y)dydx−y(1+4dydx)(x+4y)2=7x6\frac{(x+4y)\frac{dy}{dx} - y(1 + 4\frac{dy}{dx})}{(x+4y)^2} = 7x^6.

See? Not so bad, right? We've successfully differentiated both sides of the equation. Now, we have an equation that involves dydx\frac{dy}{dx}, which is exactly what we want, because this represents the slope of the tangent line. Next, we need to solve this equation for dydx\frac{dy}{dx}. This might look a little messy, but with a little algebra, we can definitely do this! Remember, practice makes perfect. The more you work through these problems, the easier they'll become. So, let's keep going and find the solution. You're doing great, and we're almost there!

Solving for dydx\frac{dy}{dx}

Okay, guys, let's isolate dydx\frac{dy}{dx}. This involves some algebraic manipulation, so let's carefully go through it step by step. First, multiply both sides of the equation by (x+4y)2(x+4y)^2 to get rid of the fraction:

(x+4y)dydx−y(1+4dydx)=7x6(x+4y)2(x+4y)\frac{dy}{dx} - y(1 + 4\frac{dy}{dx}) = 7x^6(x+4y)^2.

Next, expand the terms:

xdydx+4ydydx−y−4ydydx=7x6(x2+8xy+16y2)x\frac{dy}{dx} + 4y\frac{dy}{dx} - y - 4y\frac{dy}{dx} = 7x^6(x^2 + 8xy + 16y^2).

Notice that the terms 4ydydx4y\frac{dy}{dx} cancel out. Simplify the equation:

xdydx−y=7x6(x2+8xy+16y2)x\frac{dy}{dx} - y = 7x^6(x^2 + 8xy + 16y^2).

Now, add yy to both sides:

xdydx=7x6(x2+8xy+16y2)+yx\frac{dy}{dx} = 7x^6(x^2 + 8xy + 16y^2) + y.

Finally, divide by xx to solve for dydx\frac{dy}{dx}:

dydx=7x6(x2+8xy+16y2)+yx\frac{dy}{dx} = \frac{7x^6(x^2 + 8xy + 16y^2) + y}{x}.

Boom! We've done it! We've successfully solved for dydx\frac{dy}{dx}. Now we have a formula for the slope of the tangent line at any point (x,y)(x, y) on the curve. This is a huge accomplishment, and we're in the home stretch now. All we have to do is plug in the coordinates of the given point and calculate the numerical value of the slope. So, let's do it and find the final answer!

Finding the Slope at (1,−417)\left(1, \frac{-4}{17}\right)

We're in the final act, guys! Now that we have the general formula for dydx\frac{dy}{dx}, we can find the specific slope of the tangent line at the point (1,−417)\left(1, \frac{-4}{17}\right). This is where we substitute x=1x = 1 and y=−417y = \frac{-4}{17} into our equation:

dydx=7(1)6((1)2+8(1)(−417)+16(−417)2)+−4171\frac{dy}{dx} = \frac{7(1)^6((1)^2 + 8(1)(\frac{-4}{17}) + 16(\frac{-4}{17})^2) + \frac{-4}{17}}{1}.

Let's simplify this step by step. First, calculate the terms inside the parentheses:

(1)2+8(1)(−417)+16(−417)2=1−3217+16(16289)=1−3217+256289(1)^2 + 8(1)(\frac{-4}{17}) + 16(\frac{-4}{17})^2 = 1 - \frac{32}{17} + 16(\frac{16}{289}) = 1 - \frac{32}{17} + \frac{256}{289}.

Find a common denominator, which is 289:

289289−544289+256289=1289\frac{289}{289} - \frac{544}{289} + \frac{256}{289} = \frac{1}{289}.

Substitute this back into the equation:

dydx=7(1)(1289)−4171=7289−417\frac{dy}{dx} = \frac{7(1)(\frac{1}{289}) - \frac{4}{17}}{1} = \frac{7}{289} - \frac{4}{17}.

Again, find a common denominator, which is 289:

7289−68289=−61289\frac{7}{289} - \frac{68}{289} = \frac{-61}{289}.

Therefore, the slope of the tangent line at the point (1,−417)\left(1, \frac{-4}{17}\right) is −61289\frac{-61}{289}. Congratulations, we've found our answer! We started with a complex equation and ended up finding the slope. That's a huge victory. Keep up the great work, and continue practicing these techniques. You got this!

Conclusion

So, there you have it, folks! We've successfully used implicit differentiation to find the slope of the tangent line to the curve yx+4y=x7−5\frac{y}{x+4y} = x^7 - 5 at the point (1,−417)\left(1, \frac{-4}{17}\right). We went through the whole process, from understanding implicit differentiation to solving for dydx\frac{dy}{dx} and finally calculating the slope at the given point. I hope this was helpful. Implicit differentiation is a powerful tool in calculus, and understanding it can open the door to solving many related problems. Keep practicing, keep learning, and don't be afraid to tackle challenging problems. You're building skills that will be incredibly useful in your studies and beyond. And remember, math is just a puzzle, and it's fun to solve it together! Keep up the amazing work! If you have any questions, feel free to ask. Until next time, happy calculating!