IB Maths HL Derivatives

1. IB Mathematics HL – Derivatives, Differentiation from first principles

How can we find the first derivative function of from first principles?

Solution

IB Mathematics HL – Derivatives, Differentiation from first principles

We know that the definition of the first derivative function is given by the following formula:

So in our case, we have the following

Thus the first derivative function is

2. IB Math HL, Differentiation from first principles

IB Mathematics HL – Derivatives, Differentiation from first principles

How can we find the first derivative function of from first principles?

Solution

IB Mathematics HL – Derivatives, Differentiation from first principles

We know that the definition of the first derivative function is given by the following formula:

So, we have the following

Thus the first derivative function is

3. IB Math HL, Differentiation Product Rule

IB Mathematics HL – Derivatives, Differentiation Product Rule

How can we find the first derivative function of .

Solution

IB Mathematics HL – Derivatives, Differentiation Product Rule (or Leibnitz rule)

We know that the derivative of the product of two functions is given by the following formula

So about your question we notice that can be trated as a product of two functions .

Using the product rule we obtain

4. IB Maths HL, Differentiation Quotient Rule

IB Mathematics HL – Derivatives, Differentiation Quotient Rule

How can we find the first derivative function of .

Solution

IB Mathematics HL – Derivatives, Differentiation Quotient Rule

We know that the derivative of the quotient of two functions is given by the following formula

So about your question we notice that can be treated as a quotient of two functions .

Using the quotient rule we obtain

5. IB Maths HL, Differentiation Chain Rule

IB Mathematics HL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of ?

Solution

IB Mathematics HL – Derivatives, Differentiation Chain Rule

We know that the derivative of composite functions is given by the following formula

The chain rule states that the derivative of a function of a function (composite function) equals derivative of the outer function times derivative of the inner function.

We can view as a composite function , where .

Using the Chain rule we obtain

6. IB Maths HL, Chain Rule

IB Mathematics HL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of .

Solution

IB Mathematics HL – Derivatives, Differentiation Chain Rule

We know that the derivative of composite functions is given by the following formula

The chain rule states that the derivative of a function of a function (composite function) equals derivative of the outer function times derivative of the inner function.

So about your question we view as a composite function , where .

Using the Chain rule we obtain

7. IB Mathematics HL, Chain Rule

IB Maths HL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of

Solution

IB Mathematics HL – Derivatives, Differentiation Chain Rule

We know that the derivative of composite functions is given by the following formula

The chain rule states that the derivative of a function of a function (composite function) equals derivative of the outer function times derivative of the inner function.

So about your question we view as a composite function , where .

Using the Chain rule we obtain

8. IB Maths HL, Optimization problems

IB Mathematics HL – Derivatives, Differentiation, Optimization problems

How can we find the largest area of a rectangular region having perimeter 400 m ?

Solution

IB Mathematics HL – Derivatives, Differentiation, Optimization problems

Let be the dimensions of the rectangular region and be its area. We want to find the largest value of A.

We know that . We need to eliminate using the fact that

Thus the function described the area can be written as

Therefore the problem now is to maximize the function

To find the maximum of  , we find the stationary points

Then we use the second derivative test to determine the nature of the stationary point

So, by the second derivative test, the function has a maximum when and the largest area of a rectangular region is

and the rectangular will be a square

9. IB Math HL, Optimization problems

IB Mathematics HL – Derivatives, Differentiation, Optimization problems

How can we determine the dimensions of a cylindrical can that will hold 1.5 liliters and will minimize the amount of material used in its construction?

Solution

IB Mathematics HL – Derivatives, Differentiation, Optimization problems

Let be the radius and the height of the cylinder respectively.

Let be its area. We want to find the minimum value of A.

We know that . We need to eliminate the variable using the fact that

Thus the function described the area can be written as

Therefore the problem now is to minimize the function

To find the minimum of  , we find the stationary points

Then we use the first derivative test to determine the nature of the stationary point

It is easy to see that for all

and for all

Thus the minimum value of the area occurs when and the minimum area amount of material used in its construction is

10. IB Maths HL, Equation of Tangent

IB Mathematics HL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the tangent to the curve at the point .

Solution

IB Mathematics HL – Derivatives, Differentiation, Tangent and Normals

We know that the gradient of a curve at any point is equal to the gradient of the tangent () to the curve at that point.

So, we have that

and the gradient of the tangent () at the point is

and the equation of the tangent is given by the following formula

where in this case and finally, the equation of the tangent is

11. IB Maths HL, Equation of Normal

IB Mathematics HL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the normal to the curve at the point .

Solution

IB Mathematics HL – Derivatives, Differentiation, Tangent and Normals

We know that the gradient of a curve at any point is equal to the gradient of the tangent to the curve at that point. The slope of the normal () is the negative reciprocal of the tangent’s slope ()  since the normal is perpendicular to the tangent.

So, we have that and the gradient of the tangent () at the point is

thus the slope of the normal is

and the equation of the normal is given by the following formula

where in your case and finally, the equation of the tangent is

12. IB Maths HL, Tangent line

IB Mathematics HL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the tangent to the curve that is parallel to the line with equation .

Solution

IB Mathematics HL – Derivatives, Differentiation, Equations of Tangent and Normals

We know that the gradient of a curve at any point is equal to the gradient of the tangent () to the curve at that point.

So, we have that

and the gradient of the tangent must be equal to  since is parallel to the line

and the equation of the tangent is given by the following formula where in this case

and finally the equation of the tangent is

12. IB Maths HL, Implicit Differentiation

IB Mathematics HL – Derivatives, Implicit Differentiation

How can we find for  .

Solution

IB Mathematics HL – Derivatives, Implicit Differentiation

Sometimes functions are given not in the form but in a more complicated form in which is difficult to express y explicitly in terms of x. Such functions are called implicit functions. In these cases, we are differentiating with respect to x and for the y-terms the Chain Rule is applied.

Differentiating with respect to x:

13. IB Maths HL Implicit Differentiation

IB Mathematics HL – Derivatives, Implicit Differentiation

How can we find for  .

Solution

IB Mathematics HL –Implicit Differentiation

Sometimes functions are given not in the form but in a more complicated form in which is difficult to express y explicitly in terms of x. Such functions are called implicit functions. In these cases, we are differentiating with respect to x and for the y-terms the Chain Rule is applied.

Differentiating with respect to x:

14. IB Maths HL Stationary points

IB Mathematics HL – Derivatives, monotonicity, increasing functions, turning points

How can we find monotonicity (increasing or decreasing) and the turning points for the function  .

Solution

IB Mathematics HL – Derivatives, monotonicity, increasing functions, turning points

The first derivative function is

By setting the derivative equal to zero,

So, we know that there is a stationary point when  .

From the derivative we know that since when  the function is increasing for

Similarly, since when  the function is decreasing for .

Therefore, we can deduce that the stationary point at is a local maximum.

15. IB Maths HL Second Derivative test

IB Mathematics HL – Derivatives, Turning points, 2nd derivative test

How can we find coordinates and nature of all the stationary points of the function .

Solution

IB Mathematics HL – Derivatives, monotonicity, increasing functions, turning points

We know that

If and then at the function has a minimum turning point

If and then at the function has a maximum turning point

If and and the second derivative function changes sign around then at the function has a stationary point of inflection.

The first derivative function is

By setting the derivative equal to zero,

So, the x-coordinates of the stationary points are  .

By substituting into the original equation

and

Thus the coordinates of the stationary points are

The second derivative function is

By substituting the values of the stationary points into the second derivative function

, so is a local minimum point and

, so is a local maximum point

16. IB Maths HL Point of inflection

IB Mathematics HL – Derivatives, 2nd derivative test and Points of inflection

How can we find if the function has a point of inflexion?

Solution

IB Mathematics HL – Derivatives, Points of inflexion

We know that

If and the second derivative function changes sign around then at the function has a point of inflection.

The first derivative function is

The second derivative function is

By setting the second derivative equal to zero,

To determine the set of values for which the function is concave up or concave down we need to solve the following inequality

We observe that the function is concave up when and is concave down when

Thus at the function has a point of inflexion which has coordinates (0,16).

17. Differentiation, Derivatives, gradient, tangent- IB Mathematics HL

How can we find the equation of the tangent to the curve with equation

at the point

Solution

The gradient of the curve at is

The equation of the tangent on this point is given by the equation

, where  is the gradient of the tangent and the touch point is .

So, the equation of the tangent at the given point is

18. Differentiation, Derivatives, gradient, tangent – IB Mathematics HL

How can we find the equation of the tangent to the curve with equation

at the point ?

Solution

The gradient of the curve at is

The equation of the tangent on this point is given by the equation

where  is the gradient of the tangent and the touch point is .

So, the equation of the tangent at the given point is