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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 \( y = 6t^3 \) and \( x = 3t^2 \).

Correct Wrong

3. Find \( \frac{dy}{dx} \) when \( t = 8\) if \( y = 6t^3 + 2\) and \( x = 6t - 4\).

Correct Wrong

4. Find \( \frac{dy}{dx} \) when \( t = 9\) if \( y = 4t^3 + 2\) and \( x = 6t - 2\).

Correct Wrong

5. Find \( \frac{dy}{dx} \) when \( t = 2\) 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 \( y = 9t^2 - 3t\) and \( x = 2t - 6\) at the point where \( t = 6\).

Correct Wrong

8. Find the gradient of the tangent to the curve produced by the parametric equation \( y = t^5(2t - 5)^3\) and \( x = 10t + 2\) at the point where \( t = 1\).

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 \(y=e^{2t}+3\) and \( x = e^t - 8 \) and \( \frac{dy}{dx} = 2e\).

Correct Wrong

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 Integration


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Description of Levels



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.

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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

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