Calculus

Explanation

• For a curve $y=f(x)$, the derivative $\frac{dy}{dx}$ is its gradient function. It gives the instantaneous rate of change of $y$ with respect to $x$, rather than an average change across an interval.
• Substituting a chosen $x$-coordinate into $\frac{dy}{dx}$ gives the gradient at that point.
• A positive derivative means the curve is increasing locally, a negative derivative means it is decreasing locally, and a zero derivative identifies a stationary point candidate.
• Units of a rate are units of $y$ per unit of $x$.
• Examiners expect substitution into the derivative, not into the original function, when a rate of change is requested.

Explanation

• At a point on a differentiable curve, $f'(x)$ or $\frac{dy}{dx}$ is the gradient of the tangent at that point. The tangent is the straight line that matches the curve’s instantaneous direction locally.
• To find its gradient, differentiate the function and substitute the point’s $x$-coordinate into the derivative.
• If a gradient is prescribed, set the derivative equal to that value and solve for every possible $x$; the original function then supplies the corresponding coordinates.
• Examiners expect a distinction between the point’s $y$-coordinate and its tangent gradient.
• A curve may have the same tangent gradient at more than one point, so all valid solutions must be considered.

Explanation

• For integer $n$, the power rule is $\frac{d}{dx}(kx^n)=knx^{n-1}$. Each term in a sum or difference is differentiated separately, preserving its sign.
• A constant differentiates to $0$. Negative integer powers use the same rule, so a fraction such as $\frac{k}{x^3}$ should first be written $kx^{-3}$.
• Expressions that are products of brackets may need expanding and simplifying before term-by-term differentiation, as required by this specification.
• Examiners expect both parts of the power rule: multiply the coefficient by the old power and reduce the power by $1$.
• Final answers may be written with negative powers or converted back to fractions.

Explanation

• To find a tangent equation, first determine the point on the curve and evaluate $\frac{dy}{dx}$ there to obtain the tangent gradient. The normal is perpendicular to the tangent, so when both gradients are defined its gradient is the negative reciprocal.
• Use the common point of contact in point-gradient form $y-y_1=m(x-x_1)$.
• If only an $x$-coordinate is supplied, substitute into the original curve to find $y$ before forming either line.
• Horizontal tangents have vertical normals, which require an equation $x=\text{constant}$ rather than a finite gradient.
• Examiners expect the curve point, gradient calculation and line equation as distinct method stages.

Explanation

• A differentiable function is increasing where $\frac{dy}{dx}>0$ and decreasing where $\frac{dy}{dx}<0$. Solve $\frac{dy}{dx}=0$ to find stationary boundary values, then determine the derivative’s sign in every interval created by those values.
• This can be done by factor signs or test values.
• The sign of $y$ itself is irrelevant: a curve below the $x$-axis may still be increasing.
• At a stationary point the derivative is zero, so strict increasing and decreasing intervals normally use open endpoints.
• Examiners expect interval notation or inequalities covering every region and a derivative sign argument, not a visual judgment from an unscaled sketch.

Explanation

• The second derivative $\frac{d^2y}{dx^2}$ is the derivative of the gradient function $\frac{dy}{dx}$, so it measures how the gradient changes as $x$ changes. It is found by differentiating the original function twice.
• At a point, $\frac{d^2y}{dx^2}>0$ means the gradient is increasing and $\frac{d^2y}{dx^2}<0$ means the gradient is decreasing.
• This sign can help classify stationary points.
• The notation does not mean $\left(\frac{dy}{dx}\right)^2$; the superscript records a second differentiation.
• Examiners expect both derivative stages when the original function is supplied, with negative powers simplified consistently.

Explanation

• At a stationary point, $\frac{dy}{dx}=0$. Solve this equation and substitute each resulting $x$-value into the original curve to obtain full coordinates.
• The nature can be determined using the second derivative: a negative value gives a local maximum and a positive value gives a local minimum.
• Alternatively, a derivative sign change from positive to negative proves a maximum, while negative to positive proves a minimum.
• If the second derivative is zero, this test is inconclusive and the sign-change method is needed.
• Examiners expect every stationary point, its coordinates and its nature; reporting only the $x$-coordinates or assuming that every stationary point is a turning point is incomplete.

Explanation

• In an optimisation problem, express the quantity to be maximised or minimised as a function of one variable. Use any perimeter, length or other constraint to remove additional variables.
• Differentiate, solve the derivative equation $\frac{dQ}{dx}=0$, and verify the required nature with the second derivative, a derivative sign change or the known shape of a quadratic.
• Reject values outside the contextual domain, such as negative lengths.
• Finally substitute the stationary input back into the requested quantity: the value of $x$ is not the final answer when the question asks for a maximum area, minimum cost or another output.
• Examiners expect the model, stationary calculation, nature check and contextual conclusion.

Explanation

• A local maximum is a turning point where the curve changes from increasing to decreasing; a local minimum changes from decreasing to increasing. To sketch from known information, plot and label the turning points and any intercepts in increasing $x$-order, then join them with one smooth curve.
• Horizontal tangents should be visible at smooth maxima and minima.
• The derivative sign determines which sections rise or fall.
• A sketch need not be to scale, but relative positions, crossings and the number of turning points must match the information.
• Examiners expect no invented roots or stationary points, and an interpretation question should use intervals between the stated turning-point coordinates.