Differentiation

Explanation

• The derivative $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ is the tangent gradient and instantaneous rate of change. For $\sin x$ and $\cos x$, expand the compound angle and use $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$.
• To sketch $f'$, record where $f$ rises or falls, where its tangents are horizontal, and how steep those tangents are; the zeros of $f'$ occur at stationary points of $f$. The second derivative is the rate of change of the gradient: $f''(x)>0$ indicates a convex section and $f''(x)<0$ a concave section.
• At an inflection point, $f''$ changes sign.
• A common error is to declare an inflection point from $f''(x)=0$ alone; a sign change of concavity must be checked.
• For first principles, use $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$; expanding $(x+h)^2$ or $(x+h)^3$ and cancelling before taking the limit derives the required power result.

Explanation

• Multiply by the power and reduce it by one: $\dfrac{d}{dx}(x^n)=nx^{n-1}$ for any rational $n$, so rewrite roots and reciprocals as powers first, $\sqrt x=x^{1/2}$ and $\dfrac{3}{x^2}=3x^{-2}$, and a fraction of powers is split term by term before differentiating.
• Each term is differentiated separately and a constant disappears; there is no $+c$ in a derivative.
• The standard results are $\dfrac{d}{dx}(e^{kx})=ke^{kx}$, $\dfrac{d}{dx}(a^{kx})=k\ln a\,a^{kx}$, $\dfrac{d}{dx}(\sin kx)=k\cos kx$, $\dfrac{d}{dx}(\cos kx)=-k\sin kx$, $\dfrac{d}{dx}(\tan kx)=k\sec^2kx$, with $x$ in radians, and $\dfrac{d}{dx}(\ln x)=\dfrac1x$; the factor $k$ from the inside is the usual omission.
• A second derivative is the derivative of the first.
• An exponential model $m=Ae^{kt}$ satisfies $\dfrac{dm}{dt}=km$, so the rate is proportional to the quantity.

Explanation

• The gradient at $x=a$ is $f'(a)$; the tangent there is $y-f(a)=f'(a)(x-a)$ and the normal has gradient $-1/f'(a)$.
• Stationary points solve $f'(x)=0$, and their nature comes from the second derivative, $f''>0$ a minimum and $f''<0$ a maximum, or from the sign of $f'$ either side; a stated calculation and a reason are needed, not just the word.
• A point of inflection needs $f''(x)=0$ and a change of sign in $f''$, so evaluate $f''$ on both sides.
• A function is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$: when $f'$ is a quadratic, find its critical values algebraically, sketch or test the sign, and give the answer as inequalities in $x$.
• Optimisation questions differentiate a formula in one variable, set the derivative to zero, then justify the maximum or minimum and answer in context with units.

Explanation

• Chain rule: differentiate the outer function, keep the inside unchanged, multiply by the derivative of the inside, so $e^{4x^2-1}$ gives $8xe^{4x^2-1}$ and $(2x+1)^5$ gives $10(2x+1)^4$; writing $u=4x^2-1$ as an intermediate step is what the reports recommend. Product rule: $(uv)'=u'v+uv'$.
• Quotient rule: $\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}$, with the order of the numerator fixed and brackets kept round each product. After differentiating, factorise out the common factor, $(2x+1)^3(10x+1)$, because the factorised form is usually what the next part needs.
• Connected rates use $\dfrac{dV}{dt}=\dfrac{dV}{dh}\times\dfrac{dh}{dt}$: differentiate the formula first, then substitute the numbers.
• For an inverse function or $x$ given in terms of $y$, $\dfrac{dy}{dx}=1\Big/\dfrac{dx}{dy}$.
• The derivatives of $\sec$, $\cosec$ and $\cot$ are in the formulae booklet.

Explanation

• For an implicit relation, differentiate every term with respect to $x$ and attach a factor $\frac{\mathrm dy}{\mathrm dx}$ whenever a differentiated term contains $y$.
• For parametric equations $x=x(t)$ and $y=y(t)$, use $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$ where $\mathrm dx/\mathrm dt\ne0$.
• After finding the gradient, use the parameter or relation to obtain the actual point before writing a tangent or normal equation.
• A common error in implicit differentiation is to write $\frac{\mathrm d}{\mathrm dx}(xy)=y+x$ instead of $y+x\frac{\mathrm dy}{\mathrm dx}$.

Explanation

• Translate a rate statement into derivative notation after defining the dependent and independent variables, including their units where relevant.
• Phrases such as 'proportional to' introduce a positive constant $k$; words such as 'decreases' or 'decays' determine whether a minus sign is needed.
• A limiting or equilibrium value often appears as a difference, for example growth towards a capacity $K$ can be modelled by $\frac{\mathrm dP}{\mathrm dt}=k(K-P)$.
• A common error is to solve the differential equation when only its construction is requested, while failing to state or determine the proportionality constant.