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

• For rational $n$, $\frac{\mathrm d}{\mathrm dx}(x^n)=nx^{n-1}$ wherever the original expression and derivative are defined.
• The standard exponential results are $\frac{\mathrm d}{\mathrm dx}(e^{kx})=ke^{kx}$ and $\frac{\mathrm d}{\mathrm dx}(a^{kx})=k(\ln a)a^{kx}$.
• For angles in radians, $\frac{\mathrm d}{\mathrm dx}(\sin kx)=k\cos kx$, $\frac{\mathrm d}{\mathrm dx}(\cos kx)=-k\sin kx$, $\frac{\mathrm d}{\mathrm dx}(\tan kx)=k\sec^2kx$, and $\frac{\mathrm d}{\mathrm dx}(\ln x)=\frac1x$.
• A common error is to omit the factor $k$ created by the inner function, or the factor $\ln a$ when differentiating $a^{kx}$.
• Differentiate each term and include each inner derivative: $4(3e^{3x})+5(2\cos(2x))-3(1/x)$.

Explanation

• At $x=a$, the tangent gradient is $f'(a)$ and a non-vertical normal has gradient $-1/f'(a)$; use the point on the curve in point-gradient form. Stationary points solve $f'(x)=0$.
• Classify them by a sign change in $f'$ or, when decisive, by $f''(x)>0$ for a minimum and $f''(x)<0$ for a maximum.
• A function is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$.
• For a constrained optimisation problem, include endpoints or domain restrictions in the comparison.
• A common error is to assume every solution of $f'(x)=0$ is a maximum or minimum; a stationary point can instead be an inflection point.

Explanation

• Use $(uv)'=u'v+uv'$ for products and $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ for quotients; brackets help preserve the order in the quotient numerator. For a composite function, differentiate the outer function and multiply by the inner derivative.
• Also $\frac{\mathrm d}{\mathrm dx}(\sec x)=\sec x\tan x$, $\frac{\mathrm d}{\mathrm dx}(\cosec x)=-\cosec x\cot x$ and $\frac{\mathrm d}{\mathrm dx}(\cot x)=-\cosec^2x$.
• Connected rates use a shared variable, for example $\frac{\mathrm dV}{\mathrm dt}=\frac{\mathrm dV}{\mathrm dr}\frac{\mathrm dr}{\mathrm dt}$.
• For an inverse $g=f^{-1}$, $g'(y)=1/f'(x)$ when $y=f(x)$ and $f'(x)\ne0$.
• A common error is to substitute numerical values before differentiating, which can erase the changing relationship between the variables.

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.