IB Math HL – Differentiation

IB Maths HL Derivatives

Questions and Answers

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

How can we find the first derivative function of f(x)=x^3 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:

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)^3-x^3}{h}=

=\lim_{h\to 0}\frac{x^3+3hx^2+3xh^2+h^3-x^3}{h}=

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

=\lim_{h\to 0}(3x^2+3xh+h^2) = 3x^2

Thus the first derivative function is f'(x)= 3x^2

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 f(x)=x^4 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:

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

So, we have the following

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

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

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

=\lim_{h\to 0}[(x+h)+x][(x+h)^2+x^2] = 4x^3

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

3. IB Math HL, Differentiation Product Rule

IB Mathematics HL – Derivatives, Differentiation Product Rule

How can we find the first derivative function of h(x)=x^4 e^x .

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

[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^4 \ and \ e^x .

Using the product rule we obtain

[ x^4 e^x]'=( x^4)'(e^x)+ ( x^4) ( e^x)'=4x^3 e^x+x^4 e^x=x^3 e^x(4+x)

4. IB Maths HL, Differentiation Quotient Rule

IB Mathematics HL – Derivatives, Differentiation Quotient Rule

How can we find the first derivative function of h(x)=\frac{x^4}{e^x} .

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

 (\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^4 \ and\ e^x .

Using the quotient rule we obtain

 (\frac{x^4}{e^x})'=\frac{( x^4)'(e^x)+ ( x^4) ( e^x)'}{(e^x)^2}=

 =\frac{4x^3 e^x + x^4 e^x}{e^{2x}}=

 =\frac{ e^x (4x^3 + x^4 )}{e^{2x}}=

 =\frac{4x^3 + x^4 }{e^x}=

5. IB Maths HL, Differentiation Chain Rule

IB Mathematics HL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of h(x)=sin(4x) ?

Solution

IB Mathematics HL – 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.

We can view h(x) as a composite function f(g(x)) , where f(y)=sin(y) \ and\ g(x)=4x.

Using the Chain rule we obtain

 h'(x)=(sin(4x))'=

 =cos(4x) (4x)'=4cos(4x)

6. IB Maths HL, Chain Rule

IB Mathematics HL – Derivatives, Differentiation Chain Rule

How can we find the first derivative function of h(x)=cos(5x^2+2x) .

Solution

IB Mathematics HL – 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)= 5x^2+2x .

Using the Chain rule we obtain

 h'(x)=(cos(5x^2+2x))'=

 =-sin(5x^2+2x)(5x^2+2x)'=-sin(5x^2+2x)(10x+2)

7. IB Mathematics HL, Chain Rule

IB Maths HL – Derivatives, Differentiation Chain Rule

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

Solution

IB Mathematics HL – 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)=cos(5x^2+2x) .

Using the Chain rule we obtain

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

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

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

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

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 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=400\Rightarrow x+y=200 \Rightarrow y=200-x

Thus the function described the area can be written as

A(x)=x(200-x)=-x^2+200x

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+200=0 \Rightarrow x=100

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

A''(x)= -2

A''(100)= -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(100)= -100^2+200(100)=-10000+20000=10000 m^2

and the rectangular will be a square (x=y=100 m)

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

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

We know that A=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

A(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 A(r)

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

A'(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 A'(r)<0 for all

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

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

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 f(x)=3x^2-5x at the point (1,-2).

Solution

IB Mathematics HL – 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, we have that f' (x)=6x-5

and the gradient of the tangent (m_{T}) at the point (1,-2) is

 m_{T}=f' (1)=6-5=1

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,-2) and finally, the equation of the tangent is

y-(-2)=1(x-1) \Rightarrow y=x-3

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 f(x)=3x^2-5x at the point (1,-2).

Solution

IB Mathematics HL – 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, we have that f' (x)=6x-5 and the gradient of the tangent (m_{T}) at the point (1,-2) is  m_{T}=f' (1)=6-5=1

thus the slope of the normal is

m_{N}=-\frac{1}{ m_{T}}=-\frac{1}{ 1}=-1 and the equation of the normal is given by the following formula y-y_{0}=f'(x_{0})(x-x_{0})

where in your case (x_{0},y_{0})=(1,-2) and finally, the equation of the tangent is y-(-2)=-1(x-1) \Rightarrow y=-x-1

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 f(x)=3x^2-5x that is parallel to the line with equation y=3x-1.

Solution

IB Mathematics HL – Derivatives, Differentiation, Equations of 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, we have that

f' (x)=6x-5 and the gradient of the tangent must be equal to  3 since is parallel to the line y=3x-1

 m_{T}= f' (x)=6x-5=3\Rightarrow 6x=8\Rightarrow x=\frac{4}{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})=(\frac{4}{3},-\frac{4}{3}) and finally the equation of the tangent is y-(-\frac{4}{3})=3(x-\frac{4}{3}) \Rightarrow y=3x-\frac{16}{3}

12. IB Maths HL, Implicit Differentiation

IB Mathematics HL – Derivatives, Implicit Differentiation

How can we find \frac{dy}{dx} for  x^3y-2y=3x-1.

Solution

IB Mathematics HL – Derivatives, Implicit Differentiation

Sometimes functions are given not in the form y=f(x) 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:

\frac{d}{dx}( x^3y-2y)=\frac{d}{dx}(3x-1)

\frac{d}{dx}( x^3y)- \frac{d}{dx}(2y)=3

(\frac{d}{dx}( x^3)y+ x^3\frac{dy}{dx})-2 \frac{dy}{dx}=3

( 3x^2 y+ x^3\frac{dy}{dx}-2 \frac{dy}{dx}=3

\frac{dy}{dx}(x^3-2) =3-3x^2y

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

13. IB Maths HL Implicit Differentiation

IB Mathematics HL – Derivatives, Implicit Differentiation

How can we find \frac{dy}{dx} for  x^3y-2y^2=3x+2.

Solution

IB Mathematics HL –Implicit Differentiation

Sometimes functions are given not in the form y=f(x) 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:

\frac{d}{dx}( x^3y-2y^2)=\frac{d}{dx}(3x+2)

\frac{d}{dx}( x^3y)- \frac{d}{dx}(2y^2)=3

(\frac{d}{dx}( x^3)y+ x^3\frac{dy}{dx})-4y \frac{dy}{dx}=3

( 3x^2 y+ x^3\frac{dy}{dx}-4y \frac{dy}{dx}=3

\frac{dy}{dx}(x^3-4y) =3-3x^2y

\frac{dy}{dx}=\frac{3-3x^2y}{ x^3-4y}

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  y=-x^2+4.

Solution

IB Mathematics HL – 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 .

Therefore, we can deduce that the stationary point at  x=0 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 f(x)=x^3-12x+16.

Solution

IB Mathematics HL – 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, the x-coordinates of the stationary points are   x=2 \or\ x=-2 .

By substituting x into the original equation

f(2)=2^3-12(2)+16=0 and f(-2)=(-2)^3-12(-2)+16=32

Thus the coordinates of the stationary points are (2,0) \ and\ (-2,32)

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

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 f(x)=x^3-12x+16 has a point of inflexion?

Solution

IB Mathematics HL – 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 (0,16).

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

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

 2x^2+5xy-4y^2-22x=-60 at the point (5,0)

Solution

 2x^2+5xy-4y^2-22x=-60 \Rightarrow

\frac{d}{dx}{ 2x^2+5xy-4y^2-22x}=\frac{d}{dx}-60 \Rightarrow

4x+5(y+x \frac{dy}{dx})-8y \frac{dy}{dx}-22=0 \Rightarrow

4x+5y+5x \frac{dy}{dx}-8y \frac{dy}{dx}-22=0 \Rightarrow

 \frac{dy}{dx}(5x-8y)=22-4x-5y \Rightarrow

 \frac{dy}{dx}=\frac{22-4x-5y}{5x-8y}\Rightarrow

The gradient of the curve at (5,0) is

 \frac{dy}{dx}=\frac{22-4(5)-5(0)}{5(5)-8(0)}\Rightarrow

 \frac{dy}{dx}=\frac{2}{25}

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

y-y_{0}=m_{t}(x-x_{0})

, where   m_{t}=\frac{2}{25} is the gradient of the tangent and the touch point is (x_{0},y_{0})=(5,0).

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

y=\frac{2}{25} (x-5)=\frac{2}{25} x-\frac{2}{5}

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

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

 f(x)=x^4+5x^2+3x+4 at the point (1,13)?

Solution

 f(x)=x^4+5x^2+3x+4\Rightarrow

f'(x)=4x^3+10x+3\Rightarrow

The gradient of the curve at (1,13) is

f'(1)=4(1)^3+10(1)+3=17

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

y-y_{0}=m_{t}(x-x_{0})

where   m_{t}=17 is the gradient of the tangent and the touch point is (x_{0},y_{0})=(1,13).

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

y-13=17(x-1) \Rightarrow y=17x-4

 

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