Implicit and parametric differentiationEdexcel A-Level Maths: Revision notes

When $y$ is not given as a function of $x$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. The chain rule gives $\frac{d}{dx}\left(y^{n}\right)=ny^{n-1}\frac{dy}{dx}.$ Then collect the $\frac{dy}{dx}$ terms on one side and factorise. Example: $x^{2}+y^{2}-4x+6y=12$ gives $2x+2y\frac{dy}{dx}-4+6\frac{dy}{dx}=0$, so $\frac{dy}{dx}=\frac{2-x}{y+3}$. At $(5,1)$ the gradient is $-\frac34$.

A term such as $xy$ needs the product rule: $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}.$ Example: $x^{2}+xy+y^{2}=7$ gives $2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$. At $(1,2)$ the gradient is $-\frac45$. Stationary points need $\frac{dy}{dx}=0$, here $2x+y=0$. Substituting $y=-2x$ into the equation gives $3x^{2}=7$, so $x=\pm\frac{\sqrt{21}}{3}$.

A curve can be given as $x=f(t)$, $y=g(t)$, where $t$ is the parameter. Then $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$ Example: $x=t^{2}+1$, $y=t^{3}-3t$ gives $\frac{dy}{dx}=\frac{3t^{2}-3}{2t}$, which is $\frac94$ at $t=2$. Horizontal tangents occur where $\frac{dy}{dt}=0$ (and $\frac{dx}{dt}\neq0$); vertical tangents where $\frac{dx}{dt}=0$ (and $\frac{dy}{dt}\neq0$).

The tangent at a point has the gradient of the curve there. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal. Use $y-y_1=m(x-x_1)$. Example (implicit): at $(5,1)$ on $x^{2}+y^{2}-4x+6y=12$ the tangent gradient is $-\frac34$, so the normal gradient is $\frac43$ and the normal is $4x-3y-17=0$. Example (parametric): $x=t^{2}+1$, $y=t^{3}-3t$ at $t=2$ has point $(5,2)$ and gradient $\frac94$, so the normal is $4x+9y-38=0$. For $x=2\sin t$, $y=3\cos2t$ at $t=\frac\pi6$: $\frac{dy}{dx}=\frac{-6\sin2t}{2\cos t}=-3$ at the point $\left(1,\frac32\right)$, so the tangent is $y=-3x+\frac92$.