Differentiating Further Functions

students, welcome to a lesson on Differentiating Further Functions 📈. In earlier calculus work, you learned how to differentiate simple expressions like polynomials, exponentials, and trigonometric functions. In this lesson, the goal is to go further: handling more complicated functions, building new derivatives using rules, and connecting differentiation to real situations such as motion, growth, and optimisation.

By the end of this lesson, you should be able to:

explain the main ideas and vocabulary behind differentiating further functions,
apply the correct differentiation rule in a variety of IB-style questions,
connect differentiation with the wider ideas of calculus,
understand how these skills support modelling and problem solving,
use examples to show what the derivative tells us about a function and its graph.

A derivative is not just a symbol. It measures how fast a quantity changes. For example, if a car’s distance from home is changing, the derivative tells us the car’s speed. If a population is increasing, the derivative tells us how quickly it is growing. This idea is the heart of calculus, and it becomes especially powerful when we work with more complicated functions.

Building on the Basics

Before moving to harder examples, students, it helps to recall what differentiation means. If $y=f(x)$, then the derivative is written as $f'(x)$, $\dfrac{dy}{dx}$, or sometimes $\dfrac{d}{dx}[f(x)]$. It gives the gradient of the tangent line to the graph at a point and the instantaneous rate of change of the function.

For simple functions, the rules are familiar:

$\dfrac{d}{dx}(x^n)=nx^{n-1}$,
$\dfrac{d}{dx}(e^x)=e^x$,
$\dfrac{d}{dx}(\sin x)=\cos x$,
$\dfrac{d}{dx}(\cos x)=-\sin x$.

Differentiating further functions means combining these rules with new techniques such as the product rule, quotient rule, and chain rule. These allow you to differentiate expressions that are not just one simple term.

A common IB question might ask for the derivative of something like $y=(3x^2-1)(x^3+4)$ or $y=\sin(5x^2)$ or $y=\dfrac{x^2+1}{\sqrt{x}}$. These are all examples where more than one differentiation rule is needed. ✅

The Product Rule and Quotient Rule

When two functions are multiplied, you cannot simply differentiate each part separately and multiply the results. Instead, use the product rule:

$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Suppose $y=(x^2+1)(x^3-2)$. Let $u=x^2+1$ and $v=x^3-2$. Then $u'=2x$ and $v'=3x^2$. So

$\frac{dy}{dx} = 2x(x^3-2) + (x^2+1)(3x^2).$

You may leave your answer in this form or simplify it further.

For quotients, use the quotient rule:

$\frac{d}{dx}$$\left[$$\frac{u(x)}{v(x)}$$\right]$ = $\frac{v(x)u'(x)-u(x)v'(x)}{[v(x)]^2}$

For example, if $y=\dfrac{x^2+1}{x-3}$, then $u=x^2+1$ and $v=x-3$, giving $u'=2x$ and $v'=1$.

$\frac{dy}{dx} = \frac{(x-3)(2x)-(x^2+1)(1)}{(x-3)^2}.$

A useful exam tip is to choose the method that keeps the algebra manageable. Sometimes the quotient rule works, but rewriting the expression first may be easier. For example,

$\frac{x^2+1}{x}=x+\frac{1}{x}$

is easier to differentiate than using the quotient rule directly.

The Chain Rule and Composite Functions

One of the most important ideas in this topic is the chain rule. This is used for composite functions, where one function is inside another. If

$y=f(g(x)),$

then

$\frac{dy}{dx}=f'(g(x))\cdot g'(x).$

A good way to think of this is “differentiate the outside, then multiply by the derivative of the inside.” 🌟

For example, if

$y=(3x^2-5)^4,$

then the outside function is $u^4$ and the inside function is $u=3x^2-5$. So

$\frac{dy}{dx}=4(3x^2-5)^3\cdot 6x=24x(3x^2-5)^3.$

Another example is

$y=\sin(5x).$

The derivative is

$\frac{dy}{dx}=\cos(5x)\cdot 5=5\cos(5x).$

The chain rule appears often in IB Mathematics: Analysis and Approaches HL because it works with powers, trigonometric functions, exponentials, logarithms, and combinations of these. It is especially useful in modelling: if temperature depends on time, and a cost depends on temperature, then the chain rule helps connect their rates of change.

Differentiating Exponential, Logarithmic, and Trigonometric Functions

Further differentiation includes functions that appear constantly in applications.

For exponentials:

$\frac{d}{dx}(e^{x})=e^{x},$

and more generally,

$\frac{d}{dx}(e^{kx})=ke^{kx}.$

For logarithms:

$\frac{d}{dx}(\ln x)=\frac{1}{x}$

for $x>0$, and using the chain rule,

$\frac{d}{dx}[\ln(g(x))]=\frac{g'(x)}{g(x)}.$

For trigonometric functions:

$\frac{d}{dx}$($\sin$ x)=$\cos$ x, \qquad $\frac{d}{dx}$($\cos$ x)=-$\sin$ x,

and again the chain rule gives results like

$\frac{d}{dx}[\cos(2x^3)]=-\sin(2x^3)\cdot 6x^2.$

These rules help model many real-world processes. Exponential functions often describe growth or decay, while trigonometric functions describe repeating patterns such as sound waves, tides, and seasonal change.

Implicit Differentiation and Related Rates

Sometimes $y$ is not written explicitly as a function of $x$. Instead, $x$ and $y$ may be linked by an equation such as

$x^2+y^2=25.$

To find $\dfrac{dy}{dx}$, differentiate both sides with respect to $x$ and treat $y$ as a function of $x$:

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

Then solve for $\dfrac{dy}{dx}$:

$\frac{dy}{dx}=-\frac{x}{y}.$

This method is called implicit differentiation. It is useful when a curve cannot easily be rewritten as $y=f(x)$.

Related rates use the same idea. Imagine a balloon whose radius is increasing over time. If the volume is

$V=\frac{4}{3}\pi r^3,$

then differentiating with respect to time $t$ gives

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$

This tells us how fast the volume changes when the radius changes. Such questions test careful use of notation and interpretation, not just calculation.

Higher-Order Derivatives and What They Mean

Differentiation can continue beyond the first derivative. The second derivative is written as

$\frac{d^2y}{dx^2}$

or $f''(x)$. This tells us how the first derivative is changing.

In motion, if $s(t)$ is displacement, then

$v(t)=\frac{ds}{dt}$

is velocity and

$a(t)=\frac{d^2s}{dt^2}$

is acceleration.

Second derivatives also help describe the shape of graphs. If

$\frac{d^2y}{dx^2}>0,$

the graph is concave up, and if

$\frac{d^2y}{dx^2}<0,$

the graph is concave down. This links differentiation to sketching and curve analysis.

For example, if

$y=x^3-3x,$

then

$\frac{dy}{dx}=3x^2-3$

and

$\frac{d^2y}{dx^2}=6x.$

These derivatives help identify stationary points and the nature of the curve.

Why This Topic Matters in Calculus

Differentiating further functions is not just a separate skill. It connects to the whole calculus course. Derivatives are used to solve optimisation problems, model motion, and investigate functions. They also support integration because differentiation and integration are linked by the Fundamental Theorem of Calculus.

For instance, if a function represents velocity, its integral gives displacement. If a function represents position, its derivative gives velocity. This back-and-forth between rates of change and total accumulation is one of the central ideas in calculus.

In HL work, you are expected not only to compute derivatives but also to choose methods wisely, explain your reasoning, and interpret results in context. That means showing algebra carefully, using correct notation such as $\dfrac{dy}{dx}$ or $f'(x)$, and checking whether answers make sense in the original situation.

Conclusion

students, differentiating further functions means moving beyond basic rules and combining them with powerful methods like the product rule, quotient rule, chain rule, implicit differentiation, and higher-order derivatives. These tools let you analyse more realistic mathematical models and answer questions about change, movement, and shape. The key idea is that differentiation measures how something changes, and the more advanced the function, the more carefully you must choose and apply the right rule. Mastering these techniques gives you a strong foundation for the rest of calculus and for IB problem solving. ✅

Study Notes

The derivative tells the rate of change and the slope of a tangent line.
The product rule is $\dfrac{d}{dx}[uv]=u'v+uv'$.
The quotient rule is $\dfrac{d}{dx}\left[\dfrac{u}{v}\right]=\dfrac{vu'-uv'}{v^2}$.
The chain rule is used for composite functions: differentiate the outside, then multiply by the derivative of the inside.
Common derivatives include $\dfrac{d}{dx}(e^x)=e^x$, $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$, $\dfrac{d}{dx}(\sin x)=\cos x$, and $\dfrac{d}{dx}(\cos x)=-\sin x$.
Implicit differentiation is used when $y$ is not isolated.
Related rates connect derivatives to changing quantities in time.
The second derivative helps describe concavity, acceleration, and shape.
Differentiation supports modelling, graph analysis, optimisation, and integration in the wider calculus topic.
Always use correct notation and check that your final answer fits the context.