IB Math Studies – Differential Calculus

Questions and Answers

IB Maths Studies Differentiation

 

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)