Implicit and parametric differentiationEdexcel International A Level Maths: Revision notes

An equation such as $x^2+y^2=25$ defines $y$ implicitly in terms of $x$: $y$ is not isolated. To find $\frac{dy}{dx}$, differentiate every term with respect to $x$. Terms in $x$ alone are differentiated as normal, but a term in $y$ needs the chain rule: $\frac{d}{dx}\big(y^n\big)=ny^{n-1}\frac{dy}{dx},\qquad \frac{d}{dx}\big(\sin y\big)=\cos y\,\frac{dy}{dx},\qquad \frac{d}{dx}\big(e^{y}\big)=e^{y}\frac{dy}{dx}.$ For $x^2+y^2=25$: $2x+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{x}{y}$. The answer is usually in terms of both $x$ and $y$.

A term such as $xy$ is a product of two functions of $x$, so use the product rule: $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}$. Similarly $\frac{d}{dx}(x^2y)=2xy+x^2\frac{dy}{dx}$. After differentiating, collect every term containing $\frac{dy}{dx}$ on one side, factorise and divide. Example: $x^2+xy+y^2=7$ at $(1,2)$. Differentiating: $2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}(x+2y)=-(2x+y)$ and $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$. At $(1,2)$: $-\frac{4}{5}$. Check the point lies on the curve first: $1+2+4=7$.

A curve may be given by parametric equations $x=f(t)$, $y=g(t)$. Then the gradient is found by the chain rule: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$ Example: $x=t^2+1$, $y=t^3-3t$. $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^2-3$, so $\frac{dy}{dx}=\frac{3t^2-3}{2t}$. The tangent is parallel to the $x$-axis when $\frac{dy}{dt}=0$ (with $\frac{dx}{dt}\ne0$), and parallel to the $y$-axis when $\frac{dx}{dt}=0$ (with $\frac{dy}{dt}\ne0$). Use identities such as $\sin2t=2\sin t\cos t$ to simplify.

The tangent at $(x_1,y_1)$ has equation $y-y_1=m(x-x_1)$, where $m=\frac{dy}{dx}$ at that point. The normal is perpendicular to the tangent, so its gradient is $-\frac1m$. If $m=0$ the normal is the vertical line $x=x_1$. If the tangent is vertical, the normal is horizontal: $y=y_1$. Example: on $x=t^2+1$, $y=t^3-3t$ at $t=2$ the point is $(5,2)$ and $m=\frac94$. Tangent: $y-2=\frac94(x-5)$, i.e. $9x-4y-37=0$. Normal: gradient $-\frac49$, so $y-2=-\frac49(x-5)$, i.e. $4x+9y-38=0$.

Read the form required: 'in the form $ax+by+c=0$ with integers' means clear fractions. For 'show that' questions, write every line of the differentiation and the rearrangement; the answer is given, so the method earns the marks. When asked for 'exact' coordinates leave surds and $\pi$ in your answer. If $\frac{dy}{dx}$ is needed 'in terms of $t$', simplify by cancelling common factors, but be aware a cancelled factor can hide a point where $\frac{dx}{dt}=0$.