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Differentiation

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

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This is level 11: differentiate simple functions parametrically You can earn a trophy if you get at least 7 questions correct and you do this activity online.

Use decimals rather than fractions in answers where necessary

1. Find \( \frac{dy}{dx} \) in terms of \( t \) if \( y = 4t^2 \) and \( x = 8t \).

Correct Wrong

2. Find \( \frac{dy}{dx} \) in terms of \( t \) if \( y = 2t^3 \) and \( x = 3t^2 \).

Correct Wrong

3. Find \( \frac{dy}{dx} \) when \( t = 5\) if \( y = 6t^3 + 2\) and \( x = 6t - 9\).

Correct Wrong

4. Find \( \frac{dy}{dx} \) when \( t = 6\) if \( y = 9t^3 + 9\) and \( x = 6t - 4\).

Correct Wrong

5. Find \( \frac{dy}{dx} \) when \( t = 14\) if \( y = \frac{t^3}{3} + \frac{t^2}{2}\) and \( x = \frac{t^2}{2} + t\).

Correct Wrong

6. Find \( \frac{dy}{dx} \) when \( t = 0.1\) if \( x = 2at^2\) and \( y = 4at\) and \(a\) is a constant.

Correct Wrong

7. Find the gradient of the tangent to the curve produced by the parametric equation \( y = 4t^2 - 4t\) and \( x = 5t - 5\) at the point where \( t = 6\).

Correct Wrong

8. Find the gradient of the tangent to the curve produced by the parametric equation \( y = t^5(3t - 3)^3\) and \( x = 5t + 9\) at the point where \( t = 2\).

Correct Wrong

9. Find the gradient of the tangent to the curve produced by the parametric equation \( y = 5 \cos 2t\) and \( x = - \sin 2t\) at the point where \( t = 0\).

Correct Wrong

10. Find the value of \(t\) if \(y=e^{2t}+1\) and \( x = e^t - 5 \) and \( \frac{dy}{dx} = 2e\).

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

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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Description of Levels

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

Level 1 - Differentiate basic polynomials

Level 2 - Differentiate polynomials including negative and fractional indices

Level 3 - Calculations involving 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.

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Example

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

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

Parametric Equations

if \(x\) and \(y\) are given in terms of a third variable, the parameter, which could be \(t\), then:

$$\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$$

Answer format

There are many ways you could correctly type in the answers that have a number of terms. The software in this page should recognise most of the commonly-used formats but if you are convinced you have the correct answer but it is being shown as incorrect try typing the answer in a different format. As always, check with your teacher if you are unsure.

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.

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