# Differentiation

## Practise the technique of differentiating polynomials and other functions with this self marking exercise.

##### L 1L 2L 3L 4Level 5L 6L 7L 8L 9L 10L 11Exam-StyleDescriptionHelp

This is level 5: differentiate trigonometric functions. You can earn a trophy if you get at least 7 questions correct and you do this activity online. Type in the terms of your answer in the same order as they appeared in the question. Use the ^ key to type in a power or index and use the forward slash / to type a fraction. Press the right arrow key to end the power or fraction. Click the Help tab above for more.

 $$y=5\sin x$$ $$\frac{dy}{dx}=$$ $$y=3\cos x$$ $$\frac{dy}{dx}=$$ $$y=9x-2\cos x$$ $$\frac{dy}{dx}=$$ $$y=9\sin x - 10x^2$$ $$\frac{dy}{dx}=$$ $$y=7\tan x - 12$$ $$\frac{dy}{dx}=$$ $$y=\frac{1}{4} \tan x + 17\cos x$$ $$\frac{dy}{dx}=$$ $$y=\frac{2}{x}-2\cos x$$ $$\frac{dy}{dx}=$$ $$y=6\sin x - 12 \sqrt x$$ $$\frac{dy}{dx}=$$ $$y=5\tan x - 2x^{3.9}$$ $$\frac{dy}{dx}=$$
Check

This is Differentiation level 5. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 6 Level 7 Level 8 Level 9 Level 10 Level 11

## Instructions

Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.

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© Transum Mathematics :: This activity can be found online at:
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## Description of Levels

Close

Before beginning these exercises make sure you understand Indices really well.

You should also have been shown how to differentiate from first principles.

Level 1 - Differentiate basic polynomials

Level 2 - Differentiate polynomials including negative and fractional indices

Level 3 - Find the gradient at the given point

Level 4 - Finding tangents and normals

Level 5 - Differentiate trigonometric functions

Level 6 - Differentiate exponential and natural logarithm functions

Level 7 - Differentiate using the chain rule

Level 8 - Differentiate using the product rule

Level 9 - Differentiate using the quotient rule

Level 10 - Interpreting derivatives and second derivatives, maxima, minima and points of inflection.

Level 11 - Differentiate simple functions parametrically

Exam Style questions are in the style of IB or A-level exam paper questions and worked solutions are available for Transum subscribers.

Integration - Exercises on indefinite and definite integration of basic algebraic and trigonometric functions.

Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.

## Example

The video above is from MathMateVideos.

## Mathematical Notation

Use the ^ key to type in a power or index then the right arrow or tab key to end the power.

For example: Type 3x^2 to get 3x2.

Use the forward slash / to type a fraction then the right arrow or tab key to end the fraction.

For example: Type 1/2 to get ½.

Fractions should be given in their lowest terms.

A square root sign (if required) should be typed in as \sqrt space then press the right arrow key after typing in the last term in the square root.

## Terminology and symbols

Please note that if $$y = f(x) = x^2$$ then the first differential can be shown in any of the following ways:

$$\frac{dy}{dx} = 2x$$ $$y' = 2x$$ $$f'(x) = 2x$$

## Differentiating Trigonometric Functions

$$\frac{d}{dx} (\sin x) = \cos x$$ $$\frac{d}{dx} (\cos x) = -\sin x$$ $$\frac{d}{dx} (\tan x) = \frac{1}{\cos^2 x}$$

## Differentiating Other Functions

$$\frac{d}{dx} (e^x) = e^x$$ $$\frac{d}{dx} ( \ln x) = \frac{1}{x}$$

In the following rules, $$u$$ and $$v$$ are functions of $$x$$.

## The Product Rule

$$\text{If} \quad y = uv \quad \text{then}$$ $$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$$

## The Quotient Rule

$$\text{If} \quad y = \frac{u}{v} \quad \text{then}$$ $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

## The Chain Rule

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$