Differentiation Worked ExamplesIGCSE Maths 0580

Differentiate term by term: multiply by the power, then reduce the power by one. The constant $-4$ differentiates to $0$.

Stationary points occur where the gradient is zero. Set the derivative to $0$.

Divide through by $6$ first — it makes the factorising much easier.

So $x = 1$ or $x = 2$. Substitute each back into the original equation for $y$.

And for $x = 2$.

To classify them, differentiate again. A negative second derivative means a maximum, positive means a minimum.

Test each stationary point.

First find the $y$-coordinate of the point by substituting into the curve.

Differentiate to get the gradient function.

Substitute $x = 2$ to get the gradient at that point.

Use the straight-line form $y - y_1 = m(x - x_1)$ with the point and gradient.

Expand into the form $y = mx + c$.

Differentiate with the letters left in: $2x^{m}$ becomes $2m x^{m-1}$, $bx$ becomes $b$ and the constant $3$ becomes $0$.

Match the $x$ term with $8x^{3}$. The power gives $m - 1 = 3$, so $m = 4$; check the coefficient: $2 \times 4 = 8$ ✓.

Match the constant term: the only constant in $\frac{dy}{dx}$ comes from $bx$.

(b) A turning point is where the gradient is zero. Set the given derivative equal to $0$.

Substitute $x = 1$ into the original equation, now written as $y = 2x^{4} - 8x + 3$.

(Not asked for, but a quick check of its nature: at $x = 0$ the gradient is $-8$ and at $x = 2$ it is $56$, so the curve goes down then up — a minimum.)

Differentiate.

"Decreasing" means the gradient is negative. Set up an inequality, not an equation.

Divide by $3$ and factorise.

The critical values are $x = -1$ and $x = 3$. The curve $y = (x-3)(x+1)$ is a positive parabola, so it is below the axis between its roots.

(a) Cutting $x$ from both ends of each side leaves a square base of side $24 - 2x$. When the flaps fold up, the height of the box is $x$.

Expand the squared bracket as two brackets multiplied, then multiply by $x$.

Reorder the terms to match the given expression.

(b) Differentiate term by term. The $576x$ term gives $576$.

At the greatest volume the gradient is zero. Divide by $12$ and factorise.

Check that $x = 4$ gives a maximum: the gradient is positive just before it and negative just after it.

Substitute $x = 4$ into the volume. The base is $16\text{ cm}$ square and the height is $4\text{ cm}$.

(c) If $x = 12$, the base side is $24 - 2(12) = 0$. There would be no base, so the volume would be $0$. The cut-out squares must satisfy $0 < x < 12$.

A stationary point gives you two pieces of information: the point lies on the curve, and the gradient there is zero. Use both.

Condition (i): substitute $x = 2$, $y = -3$ into the curve.

Simplify, then divide by $2$.

Condition (ii): differentiate the curve.

Substitute $x = 2$ and set equal to zero.

Subtract the equation from step 3 from the one in step 5 to eliminate $b$.

Back-substitute into $2a + b = -8$.

Check both conditions with $y = x^{3} - 2x^{2} - 4x + 5$.

Differentiate term by term. The $2x$ gives $2$; the constant $-7$ gives $0$.

Substitute $x = -1$. Take care with the odd power: $(-1)^{3} = -1$.

Evaluate, watching the double negative in the middle term.

(a) Choose two points on the tangent that are far apart and on grid crossings, such as $(2, -2)$ and $(4, 6)$. Gradient $=$ change in $y$ $\div$ change in $x$.

(b) Differentiate term by term: $x^{2} \to 2x$, $-2x \to -2$, and the constant $-1 \to 0$.

Substitute the $x$-coordinate of $T$. This agrees with the estimate from the tangent.

(c) At a turning point the gradient is zero.

Substitute $x = 1$ into the equation of the curve to find $y$.

Check the nature: the gradient is negative just before $x = 1$ (e.g. $x = 0$ gives $-2$) and positive just after (e.g. $x = 2$ gives $+2$), so the curve goes down then up. The graph confirms a U-shape.

(a) On the $x$-axis $y = 0$. Factorise.

(b) Differentiate to get the gradient function, then evaluate it at $A$ and at $B$.

Use $y - y_{1} = m(x - x_{1})$ for the tangent at $A(1, 0)$ with gradient $-2$.

And for the tangent at $B(3, 0)$ with gradient $2$.

(c) $C$ is where the tangents meet, so solve them simultaneously. As the diagram suggests, $C$ lies on the line of symmetry $x = 2$.

(d) Take $AB$ as the base: its length is $3 - 1 = 2$. The height is the vertical distance from $C$ to the $x$-axis, which is $2$ (a length, so positive).

(a) By symmetry $P$ is $(-p, 0)$, so the width $PQ = 2p$. $R$ is on the curve directly above $Q$, so the height $QR$ is the $y$-value of the curve at $x = p$.

Multiply width by height and expand to reach the given result.

(b) Differentiate $A$ with respect to $p$.

At a maximum the gradient is zero. $p = -2$ is rejected because $p > 0$.

Substitute $p = 2$ into the formula for $A$. The rectangle is $4$ wide and $8$ high.

(c) Use the second derivative. A negative value means the gradient is decreasing, so the turning point is a maximum.

(a) Differentiate every term. The $-9x$ term differentiates to $-9$ (a constant), and the constant $2$ differentiates to $0$.

Turning points are where the gradient is zero. Divide by $3$, then factorise — you already know $x = 1$ is one root, so $(x - 1)$ must be a factor.

So $H$ is at $x = -3$, as required. Substitute into the equation of the graph, taking care with the signs.

(b) The $x^{3}$ coefficient is positive, so the graph rises from bottom left to top right: up to a maximum, down to a minimum, then up again. The turning point further left, $H(-3, 29)$, is the maximum and $G(1, -3)$ is the minimum. (Check: at $x = 0$ the gradient is $-9$, at $x = 2$ it is $+15$ — down then up around $G$.)

(c) The solutions of the equation are where the horizontal line $y = k$ crosses the graph. Sketch it through $H$ and $G$: a line strictly between the minimum value $-3$ and the maximum value $29$ crosses three times. At exactly $k = -3$ or $k = 29$ it touches a turning point, giving only two solutions, and outside that band it crosses once.

So three different solutions need $k$ strictly between the two turning values. Write the answer as an inequality.

(a) Differentiate term by term; the constant $5$ gives $0$. This is a derivative, not an expression to factorise.

(b) At a turning point the gradient is zero.

Substitute each value into the original equation.

Match each point to the sketch. $M$ is to the left of the $y$-axis and high up, so it has the negative $x$-value; $N$ is to the right and below the $x$-axis.