Differentiating Special Functions (Cambridge International IGCSE Further Maths)

Differentiating Trig Functions

How do I differentiate trig functions?

• For calculus with trigonometric functions, angles must be measured in radians

• The derivative of $y=\mathrm{sin}x$ is $\frac{dy}{dx}=\mathrm{cos}x$

• The derivative of $y=\mathrm{cos}x$ is $\frac{dy}{dx}=-\mathrm{sin}x$

• The derivative of $y=\mathrm{tan}x$ is $\frac{dy}{dx}=\mathrm{sec}^{2}x$

• The following two sets of relationships can be derived using the chain rule, but are useful to know!

• For the linear function$ax+b$, where $a$ and $b$ are constants,

  • the derivative of $y=\mathrm{sin}(ax+b)$ is $\frac{dy}{dx}=a\mathrm{cos}(ax+b)$
  • the derivative of $y=\mathrm{cos}(ax+b)$ is $\frac{dy}{dx}=-a\mathrm{sin}(ax+b)$
  • the derivative of $y=\mathrm{tan}(ax+b)$ is $\frac{dy}{dx}=a\mathrm{sec}^{2}(ax+b)$

• For the general function $f(x)$,

  • the derivative of $y=\mathrm{sin}(f(x))$ is Error converting from MathML to accessible text.
  • the derivative of $y=\mathrm{cos}(f(x))$ is $\frac{dy}{dx}=-f^{'}(x)\mathrm{sin}(f(x))$
  • the derivative of $y=\mathrm{tan}(f(x))$ is $\frac{dy}{dx}=f^{'}(x)\mathrm{sec}^{2}(f(x))$

Exam Hint

• Remember that these rules only work in radians!

Worked Example a) Find $f'(x)$ for the functions

i. $f(x)=\mathrm{sin}x$

ii.$f(x)=\mathrm{cos}2x$

ii.$f(x)=3\mathrm{sin}4x-\mathrm{cos}(2x-3)$

i.

$f'(x)=\mathrm{cos}x$

ii. Use the chain rule or remember that when $y = \cos ( a x + b )$, then $\frac{ \mathrm{d} y}{ \mathrm{d} x} = - a \sin ( a x + b )$

$f'(x)=-2\mathrm{sin}(2x)$

iii. Differentiate 'term by term'

$\mathrm{f} ' ( x ) = 3 ( 4 \cos 4 x ) - ( - 2 \sin ( 2 x - 3 ) )$

$f'(x)=12\mathrm{cos}4x+2\mathrm{sin}(2x-3)$

b)$\begin{gathered} \\ \end{gathered}$ Find the gradient of the tangent to the curve $y=\mathrm{sin}(2x+\frac{π}{6})$ at the point where $x=\frac{π}{8}$.

Gradient of tangent is equal to gradient of curve. To find the gradient, differentiate...

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = 2 \cos ( 2 x + \frac{\pi }{6} )$

... and substitute $x=\frac{π}{8}$ into the derivative

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = 2 \cos ( 2 ( \frac{\pi }{8} ) + \frac{\pi }{6} )$

Ensure that your calculator is in radians

$=\frac{\sqrt{6}-\sqrt{2}}{2}=0.517638090...$

Final answer: Answer =$\frac{\sqrt{6}-\sqrt{2}}{2}$, or 0.518 to 3 significant figures

Differentiating e^x & lnx

How do I differentiate exponentials and logarithms?

• The derivative of $y=e^{x}$ is $\frac{dy}{dx}=e^{x}$ where $x\inℝ$

• The derivative of $y=\mathrm{ln}x$ is $\frac{dy}{dx}=\frac{1}{x}$ where $x>0$

• In addition, the results below can all be found using the chain rule but are worthwhile knowing, to save time in an exam

• For the linear function $ax+b$, where $a$ and $b$ are constants,

  • the derivative of $y=e^{(ax+b)}$ is $\frac{dy}{dx}=ae^{(ax+b)}$

  • the derivative of $y=\mathrm{ln}(ax+b)$ is $\frac{dy}{dx}=\frac{a}{(ax+b)}$

    • in the special case $b=0$, $\frac{dy}{dx}=\frac{1}{x}$ ($a$'s cancel)

• For the general function $f(x)$,

  • the derivative of $y=e^{f(x)}$ is $\frac{dy}{dx}=f'(x)e^{f(x)}$
  • the derivative of $y=\mathrm{ln}(f(x))$ is $\frac{dy}{dx}=\frac{f'(x)}{f(x)}$

Exam Hint

• Remember to avoid the common mistakes:

  • the derivative of$\mathrm{ln}kx$ with respect to $x$ is$\frac{1}{x}$, NOT$\frac{k}{x}$ !
  • the derivative of $e^{kx}$ with respect to $x$ is $ke^{kx}$, NOT $kxe^{kx-1}$!

Worked Example A curve has the equation $y=e^{-3x}+2\mathrm{ln}x$.

Find the gradient of the curve at the point where $x=2$ giving your answer in the form $a+be^{c}$, where $a,b$ and $c$ are integers to be found.

Differentiate each term separately.

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = - 3 \mathrm{e}^{- 3 x} + 2 ( \frac{1}{x} )$

Substitute $x=2$ into the derivative.

$\frac{ \mathrm{d} y}{ \mathrm{d} x} = - 3 \mathrm{e}^{- 3 ( 2 )} + 2 ( 1 \frac{1}{2} )$* *

Rearrange to the required form.

$\begin{gathered} \frac{ \mathrm{d} y}{ \mathrm{d} x} & = - 3 \mathrm{e}^{- 6} + 1 = 1 - 3 \mathrm{e}^{- 6} \end{gathered}$

Do not use your calculator to evaluate this as the question asks for the answer given in this form.

$1-3e^{-6}$