Differentiating Special Functions (DP IB Maths: AA HL)

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Paul

Last updated

14 February 2024

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Differentiating Trig Functions

How do I differentiate sin, cos and tan?

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 $\frac{d y}{d x} = \sec^{2} x$
- This result can be derived using quotient rule
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 $\frac{d y}{d x} = f' (x) \sec^{2} (f (x))$
These last three results can be derived using the chain rule
For calculus with trigonometric functions angles must be measured in radians
- Ensure you know how to change the angle mode on your GDC

Examiner Tip

As soon as you see a question involving differentiation and trigonometry put your GDC into radians mode

Worked example

Find

f' (x)

for the functions

$f (x) = \sin x$
$f (x) = \cos (5 x + 1)$

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b) A curve has equation

y = \tan (6 x^{2} - \frac{π}{4})

Find the gradient of the tangent to the curve at the point where

x = \frac{\sqrt{π}}{2}

Give your answer as an exact value.

5-2-1-ib-hl-ai-aa-extraaa-we1b-soltn-

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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 $x > 0$
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 $\frac{d y}{d x} = \frac{f' (x)}{f (x)}$
The last two sets of results can be derived using the chain rule

Examiner 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 $y = e^{- 3 x + 1} + 2 \ln 5 x$ .

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

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