Implicit and parametric differentiationAQA A-Level Maths: Revision notes

Some curves, such as $x^2+y^2-6x+4y=12$, define $y$ implicitly in terms of $x$. Differentiate every term with respect to $x$. A term in $y$ is differentiated with the chain rule: $\frac{d}{dx}\left(y^n\right)=ny^{n-1}\frac{dy}{dx}.$ Then collect the $\frac{dy}{dx}$ terms on one side, factorise, and divide. Example: $2x+2y\frac{dy}{dx}-6+4\frac{dy}{dx}=0$ gives $\frac{dy}{dx}(2y+4)=6-2x$, so $\frac{dy}{dx}=\frac{3-x}{y+2}$. The answer may involve both $x$ and $y$. Only the first derivative is needed.

A term such as $xy^2$ is a product of $x$ and $y^2$, so use the product rule: $\frac{d}{dx}\left(xy^2\right)=y^2+x\cdot2y\frac{dy}{dx}.$ Example: $xy^2+x^2=6$ gives $y^2+2xy\frac{dy}{dx}+2x=0$, so $\frac{dy}{dx}=-\frac{y^2+2x}{2xy}$. At $(2,1)$ the gradient is $-\frac54$, so the tangent is $5x+4y=14$. Terms in $x$ only are differentiated as usual, and constants disappear.

A curve can be given by $x=f(t)$ and $y=g(t)$, where $t$ is a parameter. Then $\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$ Example: $x=t^2+2t$, $y=t^3-3t$ gives $\frac{dy}{dx}=\frac{3t^2-3}{2t+2}=\frac{3(t-1)(t+1)}{2(t+1)}=\frac{3(t-1)}{2}$ for $t\ne-1$. Cancel common factors only when they are non-zero, and say so. At $t=2$ the point is $(8,2)$ with gradient $\frac32$, giving the tangent $2y=3x-20$.

The tangent at a point has the gradient $\frac{dy}{dx}$ there; the normal has gradient $-\frac{1}{m}$. Use $y-y_1=m(x-x_1)$ and give integer coefficients if asked. Stationary points need $\frac{dy}{dx}=0$, so the numerator is zero. A vertical tangent needs the denominator to be zero, with the numerator non-zero. Example: $x^2+y^2-6x+4y=12$ has $\frac{dy}{dx}=\frac{3-x}{y+2}$. The tangent is vertical where $y=-2$, which gives $x^2-6x-16=0$, so the points $(8,-2)$ and $(-2,-2)$. For a parametric curve find where $\frac{dy}{dt}=0$ (horizontal tangent) or $\frac{dx}{dt}=0$ (vertical tangent).

To eliminate the parameter, rearrange to get $\cos t=\frac x4$ and $\sin t=\frac y3$ and use $\cos^2t+\sin^2t=1$ to give $\frac{x^2}{16}+\frac{y^2}{9}=1$. Differentiating this implicitly gives $\frac{dy}{dx}=-\frac{9x}{16y}$. The parametric route gives $\frac{dy}{dx}=-\frac{3\cos t}{4\sin t}$. Substituting $x=4\cos t$ and $y=3\sin t$ shows the two answers are equal. Example: where is the gradient $-1$? Put $y=\frac{9x}{16}$ in the Cartesian form to get $x=\pm\frac{16}{5}$ and the points $\left(\frac{16}{5},\frac95\right)$ and $\left(-\frac{16}{5},-\frac95\right)$.