# Thread: Implicit Differentiation....

1. ## Implicit Differentiation....

This should be fairly simple: x^2+y=x^3+y^3

Need help understanding how to work it out and what (d/dx) and (dx/dy) mean and how they help solve the problem.

2. When you differentiate y, stick a dy/dx beside it and then solve for dy/dx.

$\displaystyle x^{2}+y-x^{3}-y^{3}=0$

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

Solve for dy/dx.

3. I would like to point out that $\displaystyle \frac{dy}{dx}$ means the derivative of y with respect to x. In actuality, you solve both x and y in the exact same manner, it just doesn't look this way because $\displaystyle \frac{dx}{dx} = 1$ which is usually an invisible coefficient.

For example, the derivative of $\displaystyle y=x$ is
$\displaystyle \frac{dy}{dx} = \frac{dx}{dx}$
and $\displaystyle \frac{dx}{dx} = 1$, so $\displaystyle \frac{dy}{dx} = 1$

Another way of saying $\displaystyle \frac{dy}{dx}$ would be y prime, or $\displaystyle y\prime$.

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So for your problem:
$\displaystyle x^{2}+y=x^{3}+y^{3}$

Prep - First, you need to know the Power Rule and the Chain Rule, the Power Rule says that the derivative of $\displaystyle x^{n}$ with respect to x is equal to $\displaystyle nx^{n-1}$. So for example, $\displaystyle \frac{d}{dx} x^{5} = 5x^{4}$ This can be proven, but that's beyond the scope of this post. Don't get confused by $\displaystyle \frac{d}{dx}$ It just means the derivative of ... with respect to x, where ... is whatever it is being multiplied by, in this case $\displaystyle x^{5}$.
The Chain Rule says that the derivative of $\displaystyle f(g(x))$ is equal to $\displaystyle f\prime(g(x))g\prime(x)$. Remember that $\displaystyle f\prime(x) = \frac{d}{dx}f(x) = \frac{df(x)}{dx}$. So if $\displaystyle f(x)=x^{5}$, and $\displaystyle g(x)=x+1$, then $\displaystyle f(g(x))=(x+1)^{5}$. Now differentiate, $\displaystyle \frac{d}{dx}f(g(x))=f\prime(g(x))g(x)$ So, just fill in the equation, $\displaystyle \frac{d}{dx}(x+1)^{5} = 5(x+1)^{4}(1)$, because the derivative of $\displaystyle g(x)$ is 1.

Step1 - differentiate both sides:
$\displaystyle 2x\frac{dx}{dx} + \frac{dy}{dx} = 3x^{2}\frac{dx}{dx} + 3y^{2}\frac{dy}{dx}$

This entire step is Power Rule combined with Chain Rule, $\displaystyle y^3$ can be looked at as 2 functions, $\displaystyle f(x) = x^{3}$ and $\displaystyle g(x) = y$. So $\displaystyle f(g(x)) = y^{3}$. This means $\displaystyle f\prime(g(x)) = 3y^{2}$ and $\displaystyle g\prime(x) = dy/dx$ or $\displaystyle y\prime$

Step2 - Simplify dx/dx as 1
$\displaystyle 2x + y\prime = 3x^{2} + 3y^{2}y\prime$

step3 - Combine like terms (put any $\displaystyle y\prime$ on the same side, and anything else on the other side)
$\displaystyle y\prime - 3y^{2}y\prime = 3x^{2} - 2x$

step4 - Factor out $\displaystyle y\prime$
$\displaystyle y\prime(1 - 3y^{2}) = 3x^{2} - 2x$

step5 - Solve for $\displaystyle y\prime$ in this case, it means divide out everything it is being multiplied by.
$\displaystyle y\prime = \frac{3x^{2} - 2x}{1 - 3y^{2}}$

So anyway, unless I messed up somewhere, that's the correct process for finding the answer for this type of a question.