BASIC DERIVATIVES PRACTICE WORKSHEET

Question 1 :

What is derivative? Explain in detail

1. Answer :

Derivative measures the rate of change in one variable or quantity with respect to the change happening in another variable or quantity.

For example, let x be an independent variable and y be a dependent variable which is depending on y.

Since y is depending on x, we can write y as a function x.

y = f(x)

Y will get different values for different values of x.

For example, let y = x² + 3.

x = 1 ----> y = 1² + 3 = 4

x = 2 ----> y = 2² + 3 = 7

x = 3 ----> y = 3² + 3 = 12

x = 4 ----> y = 4² + 3 = 19

From the above working, it is clear that for different values of x, we get different values of y.

So, the value of y is getting changed according to the value of y.

Therefore, rate of change of y is with respect to the change in x.

We can find derivative of y with respect to x. That is

y' = f'(x)

Derivative can also be defined as the slope of a curve at some value of x.

Question 2 :

Find the derivative f(x) = x³.

2. Answer :

f(x) = x³

Using the power rule of derivative,

f'(x) = 3x^{3 - 1}

= 3x²

Question 3 :

Find f^-1(x), if f(x) = x³.

3. Answer :

f(x) = x²

Using the power rule of derivative,

f'(x) = 2x^{2 - 1}

= 2x¹

= 2x

Question 4 :

Derivative of a constant is zero, why?

4. Answer :

In the answer of question 1, we have seen that the derivative of y is measuring change in y with respect to the change in the other variable x.

What if y is a constant? 

If y is a constant, then there is no change. If we measure the rate of change in the quantity that has no change or it is fixed, the result will always be zero.

Because, if there is some change in y and we measure the rate of change, we will be getting some result other than zero. But, if y is fixed or it has no change, then the rate of change of y is zero.

Therefore, the derivative of any constant is zero.

Question 5 :

Explain the following situation using derivative.

A cab driver charges $5 per mile a passenger travels.

5. Answer :

If y represents the cost for travelling in the cab and x represents the miles travelled, then the rate of change of y is 5/1 or 5. That is, the value of y is getting change by 5 units for every 1 unit change in x.

More clearly, the cost of travelling in cab is increasing by $5 for the increase of every mile.

Therefore, the derivative of y with respect to x is 5.

Question 6 :

Explain the following situation using derivative.

A cab driver has a fixed package of $100 per day travelling inside the city.

6. Answer :

Here, if a passenger travels 1 mile or two miles or any number of miles inside the city in a day, he has to pay $100 per day.

Let y represent the cost for travelling in the cab and x represent the miles travelled. Then the rate of change of y is 0. Because, the value of y is fixed (or constant), that is 100.

Therefore, the derivative of y with respect to x is 0.

Question 7 :

Find f^-1(x), if f(x) = 5x³ + 3x².

7. Answer :

f(x) = 5x³ + 3x²

Using the power rule of derivative,

f'(x) = 5(3x^{3 - 1}) + 3(2x^{2 - 1})

= 5(3x²) + 3(2x¹)

= 15x² + 6x

Question 8 :

Find f^-1(x), if f(x) = x³/3.

8. Answer :

f(x) = x³/3

Using the power rule of derivative,

f'(x) = (3x^{3 - 1})/3

= (3x²)/3

= (3/3)x²

= x²

Question 9 :

Find f^-1(x), if f(x) = -7/x².

9. Answer :

f(x) = -7/x²

f(x) = -7x ^-2

Using the power rule of derivative,

f'(x) = -7(-2x ^{-2 - 1})

= -7(-2x ^-3)

= -7(-2/x ³)

= 14/x ³

Question 10 :

Find f^-1(x), if f(x) = 5x³ + 3.

10. Answer :

f(x) = 5x³ + 3

In the function above, we have two constants 5 and 3. The constant 5 is multiplied by the variable x³ and 3 is staying alone without the variable.

When we find the derivative of f(x) = 5x³ + 3, we have to keep the constant 5 as it is. Because 5 is multiplied by the variable x³. The derivative of 3 is zero, because it is not with the variable.

f(x) = 5x³ - 3

Using the power rule of derivative,

f'(x) = 5(3x^{3 - 1}) - 0

= 5(3x²)

= 15x²

Question 11 :

Find f^-1(x), if f(x) = √x.

11. Answer :

f(x) = √x

Write the square root as exponent 1/2.

f(x) = x^1/2

Use the power rule of derivative.

f'(x) = (1/2)x^{1/2 - 1}

= (1/2)x^-1/2

= 1/(2x^1/2)

= 1/(2√x)

Question 12 :

Find f^-1(x), if f(x) = x.

12. Answer :

f(x) = √x

f(x) = x

f(x) = x¹

Use the power rule of derivative.

f'(x) = 1x^{1 - 1}

= 1x⁰

= 1(1)

= 1

Question 13 :

Find the slope of the curve f(x) = 3x³ + 5 at x = -2.

13. Answer :

f(x) = x³ + 5

Using the power rule of derivative,

f'(x) = 3x^{3 - 1} + 0

f'(x) = 3x² + 0

f'(x) = 3x²

Substitute x = -2.

f'(-2) = 3(-2)²

= 3(4)

= 12

Slope of the given curve at x = -2 is 12.

Question 14 :

Find the slope of the line f(x) = 2x - 3 at x = 3.

14. Answer :

f(x) = 2x - 3

f'(x) = 2(1) - 0

f'(x) = 2

Substitute x = 3.

f'(3) = 2

In the derivative f'(x) = 2, if we substitute any value for x, we will be getting only 2 on the right side. Because there is no variable x on the right side.

From this, we can understand that the slope of a line is a constant. That is, for any value of x, the slope of a line will be same.

So, the slope of the given line is 2.

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