Can Implicit Differentiation Be Used to Solve for dy/dx?

[KingNothing]

Hi, here is my problem. I think it has something to do with me not completely understanding implicit differentiation.

I have to find \frac{dy}{dx} of x^2+5yx+y^5=8

To do this, I differentiated the x^2 as 2x then I used the product rule to differentiate 5xy into 5y + \frac{dy}{dx} * 5x. I differentiated y^5 via the chain rule into \frac{dy}{dx}*y^4. My end result was

2x + 5y + \frac{dy}{dx} * 5x + \frac{dy}{dx} * y^4 = 0

First of all, how do I solve for \frac{dy}{dx}? Is it possible? If not, where did I go wrong?

 

[whozum]

dy/dy is always 1.

 

[whozum]

Differentiate both sides with respect to x.
x^2+5y(x)*x+y^5 = 8

The derivative of y(x) is dy/dx.

2x+5y(x)+5x(dy/dx)+5y(x)^4*(dy/dx) = 0.

The chain rule is important on the end and during the product rule.

You should be able to isolate dy/dx.

 

[KingNothing]

Saying I should be able to doesn't help. Since when has that ever helped anyone? Your differentiation looks the same as mine, although it's hard to tell unless you use latex. The function is called y, not y(x). You can call it that, however paratheses in the middle of an equation tend to cause confusion (if they are denoting arguments). Also, can you explain how to differentiate 5xy to get 5y(x)^4*(dy/dx)? I keep getting 5x as a factor like I said above.

 

[whozum]

x^2 + 5yx+y^5 = 8
Derivative of a sum is the sum of the derivatives:
(d/dx)(x^2) + (d/dx)(5yx) + (d/dx) y^5 = (d/dx) 8

(d/dx) x^2 = 2x, by power rule.
(d/dx) 5xy = 5(y+x*(dy/dx)) by product rule.
(d/dx) y^5 = 5y^4*(dy/dx) Using chain rule, deriving y as if it were x, then multiplying by the derivative of y with respect to x.
(d/dx) 8 = 0 Deriving a constant.

Add the right sides of the four above steps: 2x + 5(y+x*(dy/dx)) + 5y^4*(dy/dx) = 0

Distribute the 5 in the middle term: 2x + 5y + 5*x*(dy/dx)+5y^4*(dy/dx) = 0

Subtract 2x and 5y from both sides: 5*x*(dy/dx)+5y^4*(dy/dx) = -2x-5y

Factor out dy/dx from both terms on left: (dy/dx)*(5x+5y^4) = -2x-5y

Divide both sides by (5x+5y^4): dy/dx = (-2x-5y)/(5x+5y^4)

Sorry If I was unclear earlier, I hope this helps.



** 5y^4*dy/dx comes from the last term (y^5) being differentiated using the chain rule.

 

[Nenad]

Ill put it in latex for you:

x^2 + 5yx + y^5 = 8

now, we differentiate with respect to x, on both sides:

\frac{d}{dx} (x^2 + 5yx + y^5) = \frac{d}{dx} (8)

\frac{d}{dx} (x^2) + \frac{d}{dx} (5yx) + \frac{d}{dx} (y^5) = \frac{d}{dx} (8)

2x + 5x \frac{dy}{dx} (x) + (1)5y + 5y^4 \frac{dy}{dx} = 0

note, I had to use chain rule to solve for the derrivative of the middle term.

2x + 5x \frac{dy}{dx} + 5y + 5y^4 \frac{dy}{dx} = 0

now its just matter of rearanging:

\frac{dy}{dx} 5x + 5y^4 = -2x - 5y

\frac{dy}{dx} = \frac {-2x - 5y}{5x + 5y^4}

It looks a lot better when you write it on Latex. Take a hour or two to learn the basics, it makes this forum a lot easier to work with.

Regards,

Nenad

 

[whozum]

Is there some kind of tutorial or something?

 

[KingNothing]

Yes - https://www.physicsforums.com/showthread.php?t=8997