IB Math Studies – Differential Calculus

IB Maths Studies Differentiation

Questions and Answers

1. IB Mathematical Studies SL – Derivatives

How can we find the derivative of the following functions?

$f(x)=3x^4+5x^3-8x+4$

$g(x)=5x^3+2x^2-3x^{-1}+3$

Solution

IB Mathematical Studies SL – Derivatives

$f'(x)=12x^3+15x^2-8$

$g'(x)=15x^2+4x+3x^{-2}$

2. IB Mathematical Studies SL – Differentiation

How can we find the derivative of the following functions?

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}$

$g(x)=\frac{5x^3+2x^2-3x^{-1}}{x^2}$

Solution

IB Mathematical Studies SL – Derivatives

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}=\frac{3x^4}{3x }+\frac{5x^3}{3x }-\frac{8x }{3x }+\frac{4}{3x }=$

$= x^3+\frac{5}{3} x^2-\frac{8}{3}+\frac{4}{3}x^{-1}$

So the derivative function will be

$f'(x)= 3x^2+\frac{10}{3} x-\frac{4}{3}x^{-2}$

$g(x)=\frac{5x^3+2x^2-3x^{-1}}{x^2}=\frac{5x^3}{x^2}+\frac{2x^2}{x^2}-3x^{-3}=5x+2-3x^{-3}$

So, the derivative function will be

$g'(x)=5+9x^{-4}=5+\frac{9}{x^4}$

3. IB Mathematical Studies SL – Differentiation, Gradient of the tangent

How can we find the gradient of the tangent to $f(x)=\frac{3x^4+5x^3-8x+4}{3x}$ at $x=1$ ?

Solution

IB Mathematical Studies SL – Derivatives, Gradient of the tangent

Firstly we have to simplify the function

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}=\frac{3x^4}{3x }+\frac{5x^3}{3x }-\frac{8x }{3x }+\frac{4}{3x }=$

$= x^3+\frac{5}{3} x^2-\frac{8}{3}+\frac{4}{3}x^{-1}$

Secondly, we have to calculate the derivative function

$f'(x)= 3x^2+\frac{10}{3} x-\frac{4}{3}x^{-2}$

And finally plug-in the value $x=1$ in order to find the gradient of the tangent to the curve at this point

$f'(x)= 3(1)^2+\frac{10}{3} (1)-\frac{4}{3}(1)^{-2}=$

$= 3+\frac{10}{3}-\frac{4}{3}=3+\frac{6}{3}=3+2=5$

4. IB Mathematical Studies SL – Differentiation, Equation of the tangent

How can we find the equation of the tangent to $f(x)=\frac{3x^4+5x^3-8x+4}{3x}$ at $x=1$ ?

Solution

IB Mathematical Studies SL – Derivatives, Equation of the tangent

Firstly we have to simplify the function

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}=\frac{3x^4}{3x }+\frac{5x^3}{3x }-\frac{8x }{3x }+\frac{4}{3x }=$

$= x^3+\frac{5}{3} x^2-\frac{8}{3}+\frac{4}{3}x^{-1}$

Secondly, we have to calculate the first derivative function

$f'(x)= 3x^2+\frac{10}{3} x-\frac{4}{3}x^{-2}$

Then we have to plug-in the value $x=1$ in order to find the gradient of the tangent to the curve at this point

$f'(x)= 3(1)^2+\frac{10}{3} (1)-\frac{4}{3}(1)^{-2}=$

$= 3+\frac{10}{3}-\frac{4}{3}=3+\frac{6}{3}=3+2=5$

Since $f(1)= \frac{3(1)^4+5(1)^3-8(1)+4}{3(1)}= \frac{4}{3}$ the contact point is $(1, \frac{4}{3})$ .

Therefore the equation of the tangent to the curve $f(x)$ at $x=1$ has equation:

$y-\frac{4}{3}=5(x-1) \Leftrightarrow y=5x-5+\frac{4}{3}\Leftrightarrow$

$\Leftrightarrow y=5x-\frac{11}{3}$

5. IB Mathematical Studies SL – Differentiation, Second derivative

How can we find the second derivative of the following functions?

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}$

$g(x)=\frac{5x^3+2x^2-3x^{-1}}{x^2}$

Solution

IB Mathematical Studies SL – Derivatives, Second derivative

Firstly we have to simplify the function

$f(x)=\frac{3x^4+5x^3-8x+4}{3x}=\frac{3x^4}{3x }+\frac{5x^3}{3x }-\frac{8x }{3x }+\frac{4}{3x }=$

$= x^3+\frac{5}{3} x^2-\frac{8}{3}+\frac{4}{3}x^{-1}$

So, the first derivative function is:

$f'(x)= 3x^2+\frac{10}{3} x-\frac{4}{3}x^{-2}$ and the second derivative is:

$f''(x)= 6x+\frac{10}{3} +2\frac{4}{3}x^{-3}=$

$= 6x+\frac{10}{3} +\frac{8}{3x^3}=$

Similarly,

$g(x)=\frac{5x^3+2x^2-3x^{-1}}{x^2}=\frac{5x^3}{x^2}+\frac{2x^2}{x^2}-3x^{-3}=5x+2-3x^{-3}$

The first derivative function is

$g'(x)=5+9x^{-4}=5+\frac{9}{x^4}$ and the second derivative is:

$g''(x)= -36x^{-5}$

6. IB Mathematical Studies SL – Differentiation, Monotonicity, Increasing Decreasing Function

How can we find the intervals where $f(x)$ is decreasing or increasing?

$f(x)=x^3+2x^2+4$

Solution

IB Mathematical Studies SL – Derivatives, Monotonicity, Increasing Decreasing Function

Firstly, we have to find the first derivative function

$f'(x)= 3x^2+4x=x(3x+4)$ which is positive in the intervals

$(- \infty , -\frac{4}{3})\cup (0, +\infty)$ and is negative in the interval $(-\frac{4}{3},0)$

Therefore the function is increasing for

$x \in (- \infty , -\frac{4}{3})\cup (0, +\infty)$ and is increasing for $x \in (-\frac{4}{3},0)$