IB Math SL – Differentiation

Questions and Answers

IB Maths SL Derivatives

 

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

How can we find the first derivative function of f(x)=x^2 from first principles?

Solution

IB Mathematics SL – Derivatives, Differentiation from first principles

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

f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

So in our case, we have the following

f'(x)=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}=

=\lim_{h\to 0}\frac{x^2+2hx+h^2-x^2}{h}=

=\lim_{h\to 0}\frac{h(2x+h)}{h}=

=\lim_{h\to 0}(2x+h) = 2x

Thus the first derivative function is f'(x) = 2x

 

2. IB Maths SL, Differentiation from first principles

IB Mathematics SL – Derivatives, Differentiation from first principles

How can we find the first derivative function of f(x)=3x^4 from first principles?

Solution

IB Mathematics SL – Derivatives, Differentiation from first principles

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

f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

So in our case, we have the following

f'(x)=\lim_{h\to 0}\frac{3(x+h)^4-3x^4}{h}=

=\lim_{h\to 0}3\frac{[(x+h)-x][(x+h)+x][(x+h)^2+x^2]}{h}=

=\lim_{h\to 0}3\frac{[h[(x+h)+x][(x+h)^2+x^2]}{h}=

=\lim_{h\to 0}3[(x+h)+x][(x+h)^2+x^2] = 12x^3

Thus the first derivative function is f'(x)= 4x^3

 

3. IB Math SL, Differentiation Product Rule

IB Mathematics SL – Derivatives, Differentiation Product Rule

How can we find the first derivative function of h(x)=x^5 e^x .

Solution

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

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

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

So about your question we notice that h(x) can be trated as a product of two functions  x^5\  and  \ e^x .

Using the product rule we obtain

[ x^5 e^x]'=( x^5)'(e^x)+ ( x^5) ( e^x)'=5x^4 e^x+x^5 e^x=x^4 e^x(5+x)

 

4. IB Maths SL, Quotient Rule

IB Mathematics SL – Derivatives, Differentiation Quotient Rule

How can we find the first derivative function of h(x)=\frac{x^2}{e^x} .

Solution

IB Mathematics SL – Derivatives, Differentiation Quotient Rule

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

 (\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2 (x)}

So about your question we notice that h(x) can be treated as a quotient of two functions  x^2\  and  \ e^x .

Using the quotient rule we obtain

 (\frac{x^2}{e^x})'=\frac{( x^2)'(e^x)+ ( x^2) ( e^x)'}{(e^x)^2}=

 =\frac{2x e^x + x^2 e^x}{e^{2x}}=

 =\frac{ e^x (2x + x^2 )}{e^{2x}}=

 =\frac{2x + x^2 }{e^x}=

 

5. IB Maths SL, Differentiation Chain Rule

IB Mathematics SL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of h(x)=sin(5x) .

Solution

IB Mathematics SL – Derivatives, Differentiation Chain Rule

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

[f(g(x))]'=f'(g(x))g'(x)

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 h(x) as a composite function f(g(x)) , where f(y)=sin(y)\  and  \ g(x)=5x.

Using the Chain rule we obtain

 h'(x)=(sin(5x))'=

 =cos(5x) (5x)'=5cos(5x)

6. IB Mathematics SL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of h(x)=cos(x^2+2x) .

Solution

IB Mathematics SL – Derivatives, Differentiation Chain Rule

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

[f(g(x))]'=f'(g(x))g'(x)

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 h(x) as a composite function f(g(x)) , where f(y)=cos(y) \  and  \ g(x)= x^2+2x .

Using the Chain rule we obtain

 h'(x)=(cos(x^2+2x))'=

 =-sin(x^2+2x)(x^2+2x)'=-sin(x^2+2x)(2x+2)

 

7. IB Maths SL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of h(x)=ln[sin(5x^2+2x)].

Solution

IB Mathematics SL – Derivatives, Differentiation Chain Rule

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

[f(g(x))]'=f'(g(x))g'(x)

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 h(x) as a composite function f(g(x)) , where f(y)=ln(y) \  and  \ g(x)=sin(5x^2+2x) .

Using the Chain rule we obtain

h'(x)=[ln[sin(5x^2+2x)]]'=

 =\frac{1}{sin(5x^2+2x)}(sin(5x^2+2x)'=

 =\frac{1}{sin(5x^2+2x)}cos(5x^2+2x)(10x+2)

 =\frac{1}{sin(5x^2+2x)}cos(5x^2+2x)(10x+2)

 

8. IB Mathematics SL – Derivatives, Differentiation, Optimization problems

How can we find the largest area of a rectangular region having perimeter 800 m ?

Solution

IB Maths SL – Derivatives, Differentiation, Optimization problems

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

We know that A=xy. We need to eliminate y using the fact that 2x +2y=800\Rightarrow x+y=400 \Rightarrow y=400-x

Thus the function described the area can be written as

A(x)= x(400-x)=-x^2+400x

Therefore the problem now is to maximize the function A(x)

To find the maximum of  A(x) , we find the stationary points

A'(x)= -2x+400=0 \Rightarrow x=200

Then we use the second derivative test to determine the nature of the stationary point  x=100

A''(x)= -2

A''(200)= -2<0

So, by the second derivative test, the function A(x) has a maximum when  x=100 and the largest area of a rectangular region is

A(200)= -200^2+400(200)=-40000+80000=40000 m^2 and the rectangular will be a square (x=y=200 m)

9. IB Mathematics SL – Derivatives, Differentiation, Optimization problems

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

Solution

IB Mathematics SL – Derivatives, Differentiation, Optimization problems

Let r \  and  \ h be the radius and the height of the cylinder respectively.

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

We know that C=2\pi r^2+2\pi rh. We need to eliminate the variable h using the fact that  \pi r^2 h=1500\Rightarrow h=\frac{1500}{\pi r^2}

Thus the function described the area can be written as

C(r)= 2\pi r^2+2\pi rh =2\pi r^2+2\pi r(\frac{1500}{\pi r^2})

 =2\pi r^2+\frac{3000}{r}

Therefore the problem now is to minimize the function C(r)

To find the minimum of  C(r) , we find the stationary points

C'(r)=( 2\pi r^2+\frac{3000}{r})'=4\pi r-\frac{3000}{r^2}=0 \Rightarrow

 \Rightarrow 4\pi r-\frac{3000}{r^2}=0

 \Rightarrow \frac{4\pi r^3-3000}{r^2}=0

 \Rightarrow r=(\frac{3000}{4\pi})^{\frac{1}{3}}

 \Rightarrow r=(\frac{750}{\pi})^{\frac{1}{3}}

Then we use the first derivative test to determine the nature of the stationary point  r=(\frac{750}{\pi})^{\frac{1}{3}}

It is easy to see that C'(r)<0 for all

0<r<(\frac{750}{\pi})^{\frac{1}{3}} and C'(r)>0 for all

r>(\frac{750}{\pi})^{\frac{1}{3}}

Thus the minimum value of the area occurs when  r=(\frac{750}{\pi})^{\frac{1}{3}} and the minimum area amount of material used in its construction is

C((\frac{750}{\pi})^{\frac{1}{3}})= 725 (3 s.f)

 

10. IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the tangent to the curve f(x)=x^2-5x at the point (1,-4).

Solution

IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

We know that the gradient of a curve f(x) at any point (x_{0},y_{0}) is equal to the gradient of the tangent (m_{T}) to the curve at that point.

So about your question ,we have that

f' (x)=2x-5 and the gradient of the tangent (m_{T}) at the point (1,-4) is  m_{T}=f' (1)=2-5=-3

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

y-y_{0}=f'(x_{0})(x-x_{0})

where in this case (x_{0},y_{0})=(1,-4)

and finally ,the equation of the tangent is

y-(-4)=-3(x-1) \Rightarrow y=-3x-1

 

11. IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the normal to the curve f(x)=x^2-5x at the point (1,-4).

Solution

IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

We know that the gradient of a curve f(x) at any point (x_{0},y_{0}) is equal to the gradient of the tangent to the curve at that point. The slope of the normal (m_{N}) is the negative reciprocal of the tangent’s slope (m_{T})  since the normal is perpendicular to the tangent.

m_{T}\cdot m_{N}=-1 \Rightarrow m_{N}=-\frac{1}{ m_{T}}

So about this question, we have that

f' (x)=2x-5 and the gradient of the tangent (m_{T}) at the point (1,-4) is

 m_{T}=f' (1)=2-5=-3 thus the slope of the normal is

m_{N}=-\frac{1}{ m_{T}}=-\frac{1}{-3}=\frac{1}{3} and the equation of the normal is given by the following formula

y-y_{0}=f'(x_{0})(x-x_{0})

where in this  case (x_{0},y_{0})=(1,-4) and finally, the equation of the tangent is y-(-4)= \frac{1}{3} (x-1) \Rightarrow y=\frac{1}{3} x-\frac{13}{3}

 

12. IB Mathematics SL – Derivatives, Differentiation, Equations of Tangent and Normals

How can we find the equation of the tangent to the curve f(x)= x^2-5x that is parallel to the line with equation y=3x+2.

Solution

IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

We know that the gradient of a curve f(x) at any point (x_{0},y_{0}) is equal to the gradient of the tangent (m_{T}) to the curve at that point.

So about your question, we have that

f' (x)=2x-5

and the gradient of the tangent must be equal to  3 since is parallel to the line y=3x+2

 m_{T}= f' (x)=2x-5=3\Rightarrow 2x=8\Rightarrow x=4

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

y-y_{0}=f'(x_{0})(x-x_{0})

where in this case(x_{0},y_{0})=(4,-4) and finally, the equation of the tangent is y-(-4)=3(x-4) \Rightarrow y=3x-16

13. IB Mathematics SL – Derivatives, monotonicity, increasing functions, turning points

How can we find monotonicity (increasing or decreasing) and the turning points for the function  y=-x^2+3.

Solution

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

The first derivative function is f'(x)=-2x

By setting the derivative equal to zero, f'(x)=-2x=0\Rightarrow x=0

So we know that there is a stationary point when  x=0 .

From the derivative we know that since f'(x)=-2x>0 when  x<0 the function is increasing for  x<0

Similarly, since f'(x)=-2x<0 when  x>0 the function is decreasing for  x>0 .

From the above information, we can deduce that the stationary point at  x=0 is a local maximum.

14. IB Mathematics SL – Derivatives, Turning points, 2nd derivative test

How can we find coordinates and nature of all the stationary points of the function f(x)=x^3-12x+2.

Solution

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

We know that

If f'(x_{0})=0 and f''(x_{0})>0 then at x= x_{0} the function has a minimum turning point

If f'(x_{0})=0 and f''(x_{0})<0 then at x= x_{0} the function has a maximum turning point

If f'(x_{0})=0 and f''(x_{0})=0 and the second derivative function changes sign around x_{0} then at x= x_{0} the function has a stationary point of inflection.

The first derivative function is f'(x)=3x^2-12

By setting the derivative equal to zero,

f'(x)= 3x^2=12\Rightarrow x=\pm 2

So we know that there are stationary points when   x=2 \or\ x=-2 .

By substituting x into the original equation

f(2)=2^3-12(2)+2=-14 and f(-2)=(-2)^3-12(-2)+2=18

Thus the coordinates of the stationary points are (2,-14) \  and  \ (-2,18)

The second derivative function is f''(x)=6x

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

f''(2)=6(2)=12>0 , so is a local minimum point and

f''(-2)=6(-2)=-12<0 , so is a local maximum point

15. IB Mathematics SL , 2nd derivative test, and Points of inflection

How can we find if the function f(x)=x^3-12x+2 has a point of inflexion?

Solution

IB Mathematics SL – Derivatives, Points of inflexion

We know that

If f''(x_{0})=0 and the second derivative function changes sign around x_{0} then at x= x_{0} the function has a point of inflection.

The first derivative function is f'(x)=3x^2-12

The second derivative function is f''(x)=6x

By setting the second derivative equal to zero,

f''(x)=6x=0 \Rightarrow x =0

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

f''(x)=6x>0 \Rightarrow x >0

We observe that the function is concave up when x >0 and is concave down when x <0

Thus at x =0 the function has a point of inflexion which has coordinates (0,2).

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