UK A-Level: Differentiation Techniques

Differentiation is the mathematical engine for analysing change, a tool that transforms abstract functions into precise descriptions of gradients, velocities, and growth rates. Mastering its techniques is essential not only for exam success but for understanding the dynamics in fields from physics to economics.

The Foundational Toolkit: Product, Quotient, and Chain Rules

Many functions you encounter are built by combining simpler ones. Three essential rules allow you to dismantle these combinations systematically.

The product rule is used when you need to differentiate a function that is the product of two other functions, for example, $y=x^2\sin (x)$. The rule states: if $y=u(x)v(x)$, then the derivative is given by $$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$$ A reliable way to remember this is "first times derivative of second, plus second times derivative of first." For $y=x^2\sin (x)$, let $u=x^2$ and $v=\sin (x)$. Applying the rule gives $\frac{dy}{dx}=(x^2)(\cos (x))+(\sin (x))(2x)=x^2\cos (x)+2x\sin (x)$.

When a function is one expression divided by another, you need the quotient rule. For a function in the form $y=\frac{u(x)}{v(x)}$, the derivative is $$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.$$ A common mnemonic is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared." Consider $y=\frac{e^x}{x+1}$. Here, $u=e^x$ and $v=x+1$. Applying the formula: $\frac{dy}{dx}=\frac{(x+1)(e^x)-(e^x)(1)}{(x+1{)}^2}=\frac{e^x(x+1-1)}{(x+1{)}^2}=\frac{x{e}^x}{(x+1{)}^2}$.

The most powerful of the three is the chain rule, used for differentiating a composite function—a function within a function. If $y=f(g(x))$, we can think of it as $y=f(u)$ where $u=g(x)$. The chain rule states that $$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}.$$ This means you differentiate the outer function, leaving the inner function alone, and then multiply by the derivative of the inner function. For $y=\sin (3x^2)$, the outer function is $\sin (u)$ and the inner function is $u=3x^2$. Thus, $\frac{dy}{du}=\cos (u)$ and $\frac{du}{dx}=6x$, so $\frac{dy}{dx}=\cos (3x^2)\times 6x=6x\cos (3x^2)$.

Differentiating Exponential, Logarithmic, and Trigonometric Functions

These special function families have their own base derivatives which you must know, and they are frequently combined with the chain rule.

The exponential function $e^x$ is unique because it is its own derivative: $\frac{d}{dx}(e^x)=e^x$. More generally, for $e^{f(x)}$, you apply the chain rule: $\frac{d}{dx}(e^{f(x)})=f^{\prime }(x)\cdot e^{f(x)}$. For a general exponential $a^x$, you rewrite it as $e^{x\ln a}$ to differentiate.

The natural logarithmic function, $\ln x$, differentiates to $\frac{1}{x}$ for $x>0$. Using the chain rule for $\ln (f(x))$ gives $\frac{d}{dx}[\ln (f(x))]=\frac{f^{\prime }(x)}{f(x)}$. This result is extremely useful for implicit differentiation and for simplifying complex products via logarithmic differentiation.

For trigonometric functions, you need to know these core results:

• $\frac{d}{dx}(\sin x)=\cos x$
• $\frac{d}{dx}(\cos x)=-\sin x$
• $\frac{d}{dx}(\tan x)={\sec }^{2}x$

Remember the negative sign in the derivative of $\cos x$. When the argument is a function of $x$, apply the chain rule. For example, $\frac{d}{dx}[\cos (5x)]=-\sin (5x)\times 5=-5\sin (5x)$.

Implicit Differentiation

So far, we have differentiated functions expressed explicitly as $y=f(x)$. Implicit differentiation is used when variables are intermingled, such as in the equation of a circle $x^2+y^2=25$. Here, $y$ is not isolated. The technique involves differentiating every term with respect to $x$, remembering that $y$ is itself a function of $x$ (i.e., $y(x)$). This means whenever you differentiate a term in $y$, you must apply the chain rule and multiply by $\frac{dy}{dx}$.

Let's differentiate $x^2+y^2=25$ step-by-step:

1. Differentiate $x^2$: $2x$.
2. Differentiate $y^2$: Think of this as $(y(x){)}^2$. The derivative is $2y\times \frac{dy}{dx}$.
3. Differentiate the constant $25$: $0$.

This gives: $2x+2y\frac{dy}{dx}=0$.

1. Solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx}=-2x$, so $\frac{dy}{dx}=-\frac{x}{y}$.

This method is indispensable for equations where isolating $y$ is difficult or impossible, and it is the standard technique for finding the gradient of a curve defined implicitly.

Connected Rates of Change Problems

These problems link the rate of change of one quantity to the rate of change of another via a known relationship. They are a direct application of the chain rule in a practical context. The strategy is systematic:

1. Define variables: Clearly label all quantities, noting which are changing with time.
2. Write the connecting equation: This is the geometric or physical relation (e.g., volume of a sphere: $V=\frac{4}{3}\pi r^3$).
3. Differentiate implicitly with respect to time ($t$): Apply $\frac{d}{dt}$ to both sides of the equation.
4. Substitute known values and solve: Plug in the specific instantaneous values of variables and their rates to find the unknown rate.

Example: Air is pumped into a spherical balloon at a constant rate of $100{\text{ cm}}^3/\text{s}$. How fast is the radius increasing when the radius is $10\text{ cm}$?

1. Variables: Let $V$ be volume, $r$ be radius, $t$ be time. We know $\frac{dV}{dt}=100$, and want $\frac{dr}{dt}$ when $r=10$.
2. Equation: $V=\frac{4}{3}\pi r^3$.
3. Differentiate wrt $t$: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. (Here, $\frac{d}{dt}(r^3)=3r^2\frac{dr}{dt}$).
4. Substitute: $100=4\pi (10{)}^2\frac{dr}{dt}$.
5. Solve: $100=400\pi \frac{dr}{dt}$, so $\frac{dr}{dt}=\frac{100}{400\pi }=\frac{1}{4\pi }\text{ cm/s}$.

Common Pitfalls

1. Misapplying the Product and Quotient Rules: The most frequent error is mixing up the order of terms or the sign in the quotient rule. Always write down your $u$, $v$, $\frac{du}{dx}$, and $\frac{dv}{dx}$ before substituting into the formula. For the quotient rule, carefully maintain the subtraction as "bottom d(top) minus top d(bottom)".

1. Forgetting the Chain Rule Factor: When differentiating composite functions like $\ln (3x+1)$ or ${\sin }^2(x)$, a common mistake is to stop after the outer derivative, giving $\frac{1}{3x+1}$ or $2\sin (x)$. You must multiply by the derivative of the inner function: $\frac{1}{3x+1}\times 3$ and $2\sin (x)\times \cos (x)$ respectively.

1. Algebraic Errors in Implicit Differentiation: After differentiating implicitly, solving for $\frac{dy}{dx}$ often involves algebraic manipulation. A typical mistake is mishandling fractions or incorrectly transposing terms. Move all terms containing $\frac{dy}{dx}$ to one side cleanly before factorising.

1. Incorrect Substitution in Rates Problems: Substituting numerical values too early is a major pitfall. You must differentiate first to get a general relationship like $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. Only after this step should you substitute the specific instantaneous values (like $r=10$). Substituting $r=10$ before differentiating would treat $r$ as a constant, leading to the wrong derivative of zero.

Summary

• The product rule ($\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$), quotient rule ($\frac{d}{dx}(u/v)=(u^{\prime }v-u{v}^{\prime })/v^2$), and chain rule ($dy/dx=dy/du\cdot du/dx$) are essential for breaking down complex combined functions.
• Implicit differentiation involves differentiating both sides of an equation with respect to $x$ and treating $y$ as a function of $x$, which introduces a factor of $\frac{dy}{dx}$ when differentiating $y$-terms.
• You must memorise the base derivatives for $\sin x$, $\cos x$, $\tan x$, $e^x$, and $\ln x$, and be ready to combine them with the chain rule.
• Connected rates of change problems are solved by relating variables with an equation, differentiating with respect to time, and substituting specific values to find the required rate.
• Avoid common errors by clearly labelling parts of functions, never forgetting the chain rule multiplier, and differentiating fully before substituting numerical values in applied problems.