Differentiating Special Functions (CIE IGCSE Additional Maths)

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Differentiating Trig Functions

How do I differentiate trig functions?

For calculus with trigonometric functions, angles must be measured in radians
The derivative of $y = \sin x$ is $\frac{d y}{d x} = \cos x$
The derivative of $y = \cos x$ is $\frac{d y}{d x} = - \sin x$
The derivative of $y = \tan x$ is $\frac{d y}{d x} = \sec^{2} x$
The following two sets of relationships can be derived using the chain rule, but are useful to know!
For the linear function $a x + b$ , where $a$ and $b$ are constants,
- the derivative of $y = \sin (a x + b)$ is $\frac{d y}{d x} = a \cos (a x + b)$
- the derivative of $y = \cos (a x + b)$ is $\frac{d y}{d x} = - a \sin (a x + b)$
- the derivative of $y = \tan (a x + b)$ is $\frac{d y}{d x} = a \sec^{2} (a x + b)$
For the general function $f (x)$ ,
- the derivative of $y = \sin (f (x))$ is $\frac{d y}{d x} = f^{'} (x) \cos (f (x))$
- the derivative of $y = \cos (f (x))$ is $\frac{d y}{d x} = - f^{'} (x) \sin (f (x))$
- the derivative of $y = \tan (f (x))$ is $\frac{d y}{d x} = f^{'} (x) \sec^{2} (f (x))$

Exam Tip

Remember that these rules only work in radians!

Worked example

Find

f' (x)

for the functions

f (x) = \sin x

ii.

f (x) = \cos 2 x

ii.

f (x) = 3 \sin 4 x - \cos (2 x - 3)

f' (x) = \cos x

ii. Use the chain rule or remember that when

y = \cos (a x + b)

, then

\frac{d y}{d x} = - a \sin (a x + b)

f' (x) = - 2 \sin (2 x)

iii. Differentiate 'term by term'

f' (x) = 3 (4 \cos 4 x) - (- 2 \sin (2 x - 3))

f' (x) = 12 \cos 4 x + 2 \sin (2 x - 3)

\begin{matrix}  \end{matrix}

Find the gradient of the tangent to the curve

y = \sin (2 x + \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{d y}{d x} = 2 \cos (2 x + \frac{π}{6})

... and substitute

x = \frac{π}{8}

into the derivative

\frac{d y}{d x} = 2 \cos (2 (\frac{π}{8}) + \frac{π}{6})

Ensure that your calculator is in radians

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

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 $\frac{d y}{d x} = e^{x}$ where $x \in ℝ$
The derivative of $y = \ln x$ is $\frac{d y}{d x} = \frac{1}{x}$ 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 $a x + b$ , where $a$ and $b$ are constants,
- the derivative of $y = e^{(a x + b)}$ is $\frac{d y}{d x} = a e^{(a x + b)}$
- the derivative of $y = \ln (a x + b)$ is $\frac{d y}{d x} = \frac{a}{(a x + b)}$
  - in the special case $b = 0$ , $\frac{d y}{d x} = \frac{1}{x}$ ( $a$ 's cancel)
For the general function $f (x)$ ,
- the derivative of $y = e^{f (x)}$ is $\frac{d y}{d x} = f' (x) e^{f (x)}$
- the derivative of $y = \ln (f (x))$ is $\frac{d y}{d x} = \frac{f' (x)}{f (x)}$

Exam Tip

Remember to avoid the common mistakes:
- the derivative of $\ln k x$ with respect to $x$ is $\frac{1}{x}$ , NOT $\frac{k}{x}$ !
- the derivative of $e^{k x}$ with respect to $x$ is $k e^{k x}$ , NOT $k x e^{k x - 1}$ !

Worked example

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

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

Differentiate each term separately.

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

Substitute $x = 2$ into the derivative.

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

Rearrange to the required form.

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

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

$1 - 3 e^{- 6}$

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