Implicit Differentiation
Implicit Differentiation
\(\(\frac{dy}{dx}x^2+y^2=100\)\)
Fake way to do it
- Rearrange for \(y\)
- \(y=(100-x^2)^{\frac{1}{2}}\)
- \(\frac{dy}{dx} \frac{1}{2}(100-x^2)^{-\frac{1}{2}} \times -2x\)
- \(\frac{dy}{dx}=-\frac{x}{\sqrt{ (100-x^2) }}\)
The real way to do it
- Differentiate from left to right
- \(x^2+y^2=100\)
- \(2x+2y \frac{dy}{dx}=0\)
- \(2y \frac{dy}{dx}=2x\)
- \(\frac{dy}{dx}=\frac{2x}{-2y}\)
- \(\frac{dy}{dx}=-\frac{x}{y}\)
\(y^2+3x+xy=0\) with respect to \(x\)
- \(3y^2 \frac{dy}{dx}+3+\frac{dy}{dx}=0\)
- \(3y^2 \frac{dy}{dx}+\frac{dy}{dx}=-3\)
- \(3y^2 \frac{dy}{dx}=-3-\frac{dy}{dx}\)
\(x^3+x+y^3+3y=6\)
- \(3x^2+1+3y^2\frac{dy}{dx}+3\frac{dy}{dx}=0\)
- \(3y^2\frac{dy}{dx}+3\frac{dy}{dx}=-3x^2-1\)
- \(\frac{dy}{dx}(3x^2+3)=-3x^2-1\)
- \(\frac{dy}{dx}=-\frac{3x^2-1}{3x^2+3}\)
\(x^2+y^2+10x+2y-4xy=10\)
- \(\frac{dy}{dx}\to2x+2y \frac{dy}{dx}+10+\frac{dy}{dx}-4x\frac{dy}{dx}-4y=0\)
- \(2x+10-4y=-2y\frac{dy}{dx}-\frac{dy}{dx}+4x\frac{dy}{dx}\)
- \(\frac{dy}{dx}(-2y-1+4x)=2x+10-4y\)
- \(\frac{dy}{dx}=\frac{2x+10-4y}{-2y-1+4x}\)