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> Differentiation - part 1 (Unit 1)

Maths

Calculus

Differentiation - part 1 (Unit 1) - Test Bite

1.1.

Find for $f(x) = 2x^{4} - 3x^{3} - 7x +2$

$f'(x) = 4x^{3} - 3x^{2} - 7$

$f'(x) = 6x^{3} - 6x^{2} - 7$

$f'(x) = 8x^{3} - 9x^{2} - 7$

$f'(x) = 8x^{3} + 9x^{2} - 7$

1.2.

Find for $f(x) = x^{-7}$

$f'(x) = -7x^{6}$

$f'(x) = 7x^{-8}$

$f'(x) = -7x^{-8}$

2.1.

Find f'(x) for f(x) = (x + 1)(x - 4)

2x - 3

2x + 3

x² - 3x

2.2.

Find the derivative of f(x) = (x - 3)³

f'(x) = 3(x - 3)

f'(x) = 3x² - 18x - 27

2x - 6

3x² - 9x + 27

2.3.

Differentiate y with respect to x: y = (x² - 4)(x + 2)

3x² + 4x + 4

x² + 4x - 4

3x² + 4x - 4

3.1.

Find $f'(x) : f(x) = {6 \over x^2}$

$12 \over x^3$

$-{12 \over x^3}$

3.2.

Find the derivative of $f(x)= {{3 \surd {x^3}} \over 2}$

${9 \surd {x} }\over 2$

${9 \surd {x} }\over 8$

${9 \surd {x} }\over 4$

${3 \surd {x} }\over 4$

3.3.

Find the derivative of $y = {7 \over {4 \sqrt [3] {x^2}}$

$-7 \over {6 \sqrt [3] {x^5}$

$7 \over {6 \sqrt [3] {x^5}$

$7 \over {4 \sqrt [3] {x^5}$

$-7 \over { \sqrt [5] {x^3}$

4.1.

Find ${f'(x) : f(x)} = {x - 7 \over x}$

$f'(x) = {- 7 \over x^2}$

$f'(x) = {7 \over x}$

$f'(x) = {7 \over x^2}$

4.2.

Find the derivative of $f(x) = {{3x^{2} - 2x +1} \over x}$

$f'(x) = {{3x^{2} - 1} \over x}$

$f'(x) = {{3x^{2} - 1} \over x^2}$

4.3.

Find the derivative of $y = {{{x^2} - x - 12} \over {{x^2} - 4x}}$

${dy \over dx} = {{{2x - 1} \over {2x - 4}}}$

${dy \over dx} = {1 + 6x}$

${dy \over dx} = -\frac{-3}{x^2}$

${dy \over dx} = -\frac{-3}{x}$

Given f(x) = 3x² - 8x + 1, find the gradient of the tangent at the point on the function where x = ½.

$3\frac{3}{4}$

-5

$-2\frac{1}{4}$

Find the equation of the tangent to the function below at x = $\frac{1}{2}$

$f(x) = {x + 1\over x}$

y + 4x = 5

y = 4x + 5

y + 4x = $2\frac{3}{4}$

8y + 2x = 25

7.1.

State whether the following functions are increasing, decreasing, or stationary at the given points.

f(x) = x² + 3x, at x = -2 and at x = 2

At x = -2

(the options are)

7.2.

At x = 2

(the options are)

8.1.

For the function:

$f(x) = {x^3 \over 3} - 2x^2 + 3x -7$

Determine the position and nature of the stationary points and select the correct option from the dropdown box.

There is a maximum stationary point at x =

(the options are)

8.2.

At x = 3, there is a

(the options are)

Given y = x(x-5)², find the turning points and determine their nature.

Minimum at $x = {5 \over 3}$ , maximum at x = 5

Minimum at x = -5, maximum at $x = {3 \over 5}$

Minimum at x = 5, maximum at $x = {5 \over 3}$

Minimum at $x = {3 \over 5}$ , maximum at x = -5

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Differentiation - part 1 (Unit 1) - Test Bite

SQA - Maths

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