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

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