Differentiating Further Functions (DP IB Maths: AA HL)

Revision Note

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Maths

This Revision Note focuses on the results and derivations of results involving the less common trigonometric, exponential and logarithmic functions. As with any function, questions may go on to ask about gradients, tangents, normals and stationary points.

Differentiating Reciprocal Trigonometric Functions

What are the reciprocal trigonometric functions?

Secant, cosecant and cotangent and abbreviated and defined as

$\sec x = \frac{1}{\cos x}$ $cosec x = \frac{1}{\sin x}$ $\cot x = \frac{1}{\tan x}$

Remember that for calculus, angles need to be measured in radians
- $θ$ may be used instead of $x$
$cosec x$ is sometimes further abbreviated to $\csc x$

What are the derivatives of the reciprocal trigonometric functions?

$f (x) = \sec x$
- $f' (x) = \sec x \tan x$
$f (x) = cosec x$
- $f' (x) = - cosec x \cot x$
$f (x) = \cot x$
- $f' (x) = - {cosec}^{2} x$
These are given in the formula booklet

How do I show or prove the derivatives of the reciprocal trigonometric functions?

For $y = \sec x$
- Rewrite, $y = \frac{1}{\cos x}$
- Use quotient rule, $\frac{d y}{d x} = \frac{\cos x (0) - (1) (- \sin x)}{\cos^{2} x}$
- Rearrange, $\frac{d y}{d x} = \frac{\sin x}{\cos^{2} x}$
- Separate, $\frac{d y}{d x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}$
- Rewrite, $\frac{d y}{d x} = \sec x \tan x$
Similarly, for $y = cosec x$
- $y = \frac{1}{\sin x}$
- $\frac{d y}{d x} = \frac{\sin x (0) - (1) \cos x}{\sin^{2} x}$
- $\frac{d y}{d x} = \frac{- \cos x}{\sin^{2} x}$
- $\frac{d y}{d x} = - \frac{1}{\sin x} \times \frac{\cos x}{\sin x}$
- $\frac{d y}{d x} = - cosec x \cot x$

What do the derivatives of reciprocal trig look like with a linear functions of x?

For linear functions of the form ax+b
- $f (x) = \sec (a x + b)$
  - $f' (x) = a \sec (a x + b) \tan (a x + b)$
- $f (x) = cosec (a x + b)$
  - $f' (x) = - a cosec (a x + b) \cot (a x + b)$
- $f (x) = \cot (a x + b)$
  - $f' (x) = - a {cosec}^{2} (a x + b)$
- These are not given in the formula booklet
  - they can be derived from chain rule
  - they are not essential to remember

Exam Tip

Even if you think you have remembered these derivatives, always use the formula booklet to double check
- those squares and negatives are easy to get muddled up!
Where two trig functions are involved in the derivative be careful with the angle multiple; $x, 2 x, 3 x$ , etc
- An example of a common mistake is differentiating $y = c o s e c 3 x$
  - $\frac{d y}{d x} = - 3 c o s e c x c o t 3 x$ instead of $\frac{d y}{d x} = - 3 c o s e c 3 x c o t 3 x$

Worked example

Curve C has equation $y = 2 \cot (3 x - \frac{π}{8})$ .

Show that the derivative of

\cot x

- {cosec}^{2} x

5-8-3-ib-hl-aa-only-we1a-soltn

b) Find $\frac{d y}{d x}$ for curve C.

5-8-3-ib-hl-aa-only-we1b-soltn

c) Find the gradient of curve C at the point where $x = \frac{7 π}{24}$ .

5-8-3-ib-hl-aa-only-we1c-soltn

Differentiating Inverse Trigonometric Functions

What are the inverse trigonometric functions?

arcsin, arccos and arctan are functions defined as the inverse functions of sine, cosine and tangent respectively

$\arcsin (\frac{\sqrt{3}}{2}) = \frac{π}{3}$ which is equivalent to $\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$
$\arctan (- 1) = \frac{3 π}{4}$ which is equivalent to $\tan (\frac{3 π}{4}) = - 1$

What are the derivatives of the inverse trigonometric functions?

$f (x) = \arcsin x$
- $f' (x) = \frac{1}{\sqrt{1 - x^{2}}}$
$f (x) = \arccos x$
- $f' (x) = - \frac{1}{\sqrt{1 - x^{2}}}$
$f (x) = \arctan x$
- $f' (x) = \frac{1}{1 + x^{2}}$
Unlike other derivatives these look completely unrelated at first
- their derivation involves use of the identity $\cos^{2} x + \sin^{2} x \equiv 1$
- hence the squares and square roots!
All three are given in the formula booklet
Note with the derivative of $\arctan x$ that $(1 + x^{2})$ is the same as $(x^{2} + 1)$

How do I show or prove the derivatives of the inverse trigonometric functions?

For $y = \arcsin x$
- Rewrite, $\sin y = x$
- Differentiate implicitly, $\cos y \frac{d y}{d x} = 1$
- Rearrange, $\frac{d y}{d x} = \frac{1}{\cos y}$
- Using the identity $\cos^{2} y \equiv 1 - \sin^{2} y$ rewrite, $\frac{d y}{d x} = \frac{1}{\sqrt{1 - \sin^{2} y}}$
- Since, $\sin y = x$ , $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}$
Similarly, for $y = \arccos x$
- $\cos y = x$
- $- \sin y \frac{d y}{d x} = 1$
- $\frac{d y}{d x} = - \frac{1}{\sin y}$
- $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - \cos^{2} y}}$
- $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$
Notice how the derivative of $y = \arcsin x$ is positive but is negative for $y = \arccos x$
- This subtle but crucial difference can be seen in their graphs
  - $y = \arcsin x$ has a positive gradient for all values of $x$ in its domain
  - $y = \arccos x$ has a negative gradient for all values of $x$ in its domain

What do the derivative of inverse trig look like with a linear function of x?

For linear functions of the form $a x + b$
$f (x) = \arcsin (a x + b)$
- $f' (x) = \frac{a}{\sqrt{1 - {(a x + b)}^{2}}}$
$f (x) = \arccos (a x + b)$
- $f' (x) = \frac{a}{\sqrt{1 - {(a x + b)}^{2}}}$
$f (x) = \arctan (a x + b)$
- $f' (x) = \frac{a}{1 + {(a x + b)}^{2}}$
These are not in the formula booklet
- they can be derived from chain rule
- they are not essential to remember
- they are not commonly used

Exam Tip

For $f (x) = \arctan x$ the terms on the denominator can be reversed (as they are being added rather than subtracted)
- $f' (x) = \frac{1}{1 + x^{2}} = \frac{1}{x^{2} + 1}$
- Don't be fooled by this, it sounds obvious but on awkward "show that" questions it can be off-putting!

Worked example

a) Show that the derivative of $\arctan x$ is $\frac{1}{1 + x^{2}}$

5-8-3-ib-hl-aa-only-we2a-soltn

Find the derivative of

\arctan (5 x^{3} - 2 x)

5-8-3-ib-hl-aa-only-we2b-soltn

Differentiating Exponential & Logarithmic Functions

What are exponential and logarithmic functions?

Exponential functions have term(s) where the variable ( $x$ ) is the power (exponent)

In general, these would be of the form $y = a^{x}$

The special case of this is when $a = e$ , i.e. $y = e^{x}$

Logarithmic functions have term(s) where the logarithms of the variable ( $x$ ) are involved

In general, these would be of the form $y = \log_{a} x$

The special case of this is when $a = e$ , i.e. $y = \log_{e} x = \ln x$

What are the derivatives of exponential functions?

The first two results, of the special cases above, have been met before
- $f (x) = e^{x}, f' (x) = e^{x}$
- $f (x) = \ln x, f' (x) = \frac{1}{x}$
- These are given in the formula booklet
For the general forms of exponentials and logarithms
- $f (x) = a^{x}$
  - $f' (x) = a^{x} (\ln a)$
- $f (x) = \log_{a} x$
  - $f' (x) = \frac{1}{x \ln a}$
- These are also given in the formula booklet

How do I show or prove the derivatives of exponential and logarithmic functions?

For $y = a^{x}$
- Take natural logarithms of both sides, $\ln y = x \ln a$
- Use the laws of logarithms, $\ln y = x \ln a$
- Differentiate, implicitly, $\frac{1}{y} \frac{d y}{d x} = \ln a$
- Rearrange, $\frac{d y}{d x} = y \ln a$
- Substitute for $y$ , $\frac{d y}{d x} = a^{x} \ln a$
For $y = \log_{a} x$
- Rewrite, $x = a^{y}$
- Differentiate $x$ with respect to $y$ , using the above result, $\frac{d x}{d y} = a^{y} \ln a$
- Using $\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}$ , $\frac{d y}{d x} = \frac{1}{a^{y} \ln a}$
- Substitute for $y$ , $\frac{d y}{d x} = \frac{1}{a^{\log_{a} x} \ln a}$
- Simplify, $\frac{d y}{d x} = \frac{1}{x \ln a}$

What do the derivatives of exponentials and logarithms look like with a linear functions of x?

For linear functions of the form $p x + q$
- $f (x) = a^{p x + q}$
  - $f' (x) = p a^{p x + q} (\ln a)$
- $f (x) = \log_{a} (p x + q)$
  - $f' (x) = \frac{p}{(p x + q) \ln a}$
- These are not in the formula booklet
  - they can be derived from chain rule
  - they are not essential to remember

Exam Tip

For questions that require the derivative in a particular format, you may need to use the laws of logarithms
- With ln appearing in denominators be careful with the division law
  - $\ln (\frac{a}{b}) = \ln a - \ln b$
  - but $\frac{\ln a}{\ln b}$ cannot be simplified (unless there is some numerical connection between $a$ and $b$ )

Worked example

Find the derivative of $a^{3 x - 2}$ .

Chain rule or ' $p x + q$ shortcut' is required

$\frac{d}{d x} [a^{3 x - 2}] = a^{3 x - 2} \ln a \times 3$

$∴$ The derivative of $a^{3 x - 2}$ is $3 a^{3 x - 2} \ln a$

b) Find an expression for $\frac{d y}{d x}$ given that $y = \log_{5} (2 x^{3})$

Chain rule is needed

\frac{d y}{d x} = \frac{1}{2 x^{3} \ln 5} \times 6 x^{2}

Simplify by cancelling

∴ \frac{d y}{d x} = \frac{3}{x \ln 5}

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Differentiating Further Functions (DP IB Maths: AA HL)

Revision Note

What are the reciprocal trigonometric functions?

What are the derivatives of the reciprocal trigonometric functions?

How do I show or prove the derivatives of the reciprocal trigonometric functions?

What do the derivatives of reciprocal trig look like with a linear functions of x?

What are the inverse trigonometric functions?

What are the derivatives of the inverse trigonometric functions?

How do I show or prove the derivatives of the inverse trigonometric functions?

What do the derivative of inverse trig look like with a linear function of x?

What are exponential and logarithmic functions?

What are the derivatives of exponential functions?

How do I show or prove the derivatives of exponential and logarithmic functions?

What do the derivatives of exponentials and logarithms look like with a linear functions of x?

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Author: Paul