IB Math SL - Differentiation - IB Maths Apps & Books

IB Mathematics SL – Derivatives

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

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

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

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

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.