5.9 Differentiation rules and related ratesIB Maths: Applications and Interpretation HL: Revision notes

Know these results, where $n\in\mathbb{Q}$ (any fraction or negative power):

• $\frac{d}{dx}x^n=nx^{n-1}$, so $\frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}$ and $\frac{d}{dx}\frac1x=-\frac{1}{x^2}$
• $\frac{d}{dx}\sin x=\cos x$ and $\frac{d}{dx}\cos x=-\sin x$ (angles in radians)
• $\frac{d}{dx}\tan x=\frac{1}{\cos^2x}$
• $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}\ln x=\frac1x$ Rewrite roots and fractions as powers first. For example $\frac{3}{\sqrt{x}}=3x^{-1/2}$, with derivative $-\frac32x^{-3/2}$. Make sure your GDC is in radian mode when evaluating trigonometric derivatives.

For a composite function $y=f(g(x))$, let $u=g(x)$. Then $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.$ Differentiate the outer function, then multiply by the derivative of the inner function.

• $y=\sin(2x+5)\Rightarrow\frac{dy}{dx}=2\cos(2x+5)$
• $y=e^{-0.5t}\Rightarrow\frac{dy}{dt}=-0.5e^{-0.5t}$
• $y=\ln(3x+2)\Rightarrow\frac{dy}{dx}=\frac{3}{3x+2}$
• $y=(x^2+1)^{1/2}\Rightarrow\frac{dy}{dx}=\frac12(x^2+1)^{-1/2}(2x)=\frac{x}{\sqrt{x^2+1}}$

If $y=uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$ Example: $y=x^2\sin x$. With $u=x^2$ and $v=\sin x$: $\frac{dy}{dx}=x^2\cos x+2x\sin x$. Example with a fractional power: $y=\sqrt{t}\,e^{-0.5t}$ gives $\frac{dy}{dt}=\frac12t^{-1/2}e^{-0.5t}-0.5\sqrt t\,e^{-0.5t}=\frac{e^{-0.5t}(1-t)}{2\sqrt t}$, after factorising. Factorising makes it easy to solve $\frac{dy}{dt}=0$.

If $y=\dfrac{u}{v}$, then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.$ Example: $y=\dfrac{20t}{t^2+4}$ with $u=20t$ and $v=t^2+4$: $\frac{dy}{dt}=\frac{20(t^2+4)-20t(2t)}{(t^2+4)^2}=\frac{20(4-t^2)}{(t^2+4)^2}$. The order in the numerator matters: $v\,u'-u\,v'$, not the other way round. You can also check by writing $\tan x=\frac{\sin x}{\cos x}$ and confirming $\frac{d}{dx}\tan x=\frac{1}{\cos^2x}$.

When two quantities are linked by a formula and both change with time, use the chain rule to connect their rates: $\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}.$ Method: write the formula linking the variables, differentiate with respect to the variable you know the rate of, substitute known values, then solve. Example: a circular ripple has radius $r$ cm increasing at $2$ cm s$^{-1}$. Area $A=\pi r^2$, so $\frac{dA}{dr}=2\pi r$ and $\frac{dA}{dt}=2\pi r\times2$. When $r=5$: $\frac{dA}{dt}=20\pi=62.8$ cm$^2$ s$^{-1}$. If a volume is given, such as $V=\frac43\pi r^3$ with $\frac{dV}{dt}=50$, solve for $\frac{dr}{dt}=\frac{50}{4\pi r^2}$.

A stationary point occurs where $\frac{dy}{dx}=0$. Distinguish a maximum from a minimum by the sign of the gradient either side, or the sign of the second derivative. In context, a maximum of a model, such as the greatest concentration or highest profit, is found by solving $f'(x)=0$ and then evaluating $f$ there. For optimisation questions: define the quantity, write it as a function of one variable (using a constraint if needed), differentiate with the rules above, solve $f'(x)=0$, justify that the point is a maximum or minimum, and give the answer in context with units.