Differentiate from first principles the function $f\left(x\right)=3x^3-x$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$=\frac{\left(3{\left(x+h\right)}^3-\left(x+h\right)\right)-\left(3x^3-x\right)}{h}$    M1

$=\frac{3\left(x^3+3x^{2}h+3x{h}^2+h^3\right)-x-h-3x^3+x}{h}$     (A1)

$=\frac{9x^{2}h+9x{h}^2+3h^3-h}{h}$      A1

cancelling $h$      M1

$=9x^2+9xh+3h^2-1$

then $\underset{h\to 0}{\text{lim}}\left(9x^2+9xh+3h^2-1\right)$

$=9x^2-1$      A1

Note: Final A1 dependent on all previous marks.

 

METHOD 2

$\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$=\frac{\left(3{\left(x+h\right)}^3-\left(x+h\right)\right)-\left(3x^3-x\right)}{h}$    M1

$=\frac{3\left({\left(x+h\right)}^3-x^3\right)+\left(x-\left(x+h\right)\right)}{h}$       (A1)

$=\frac{3h\left({\left(x+h\right)}^2+x\left(x+h\right)+x^2\right)-h}{h}$      A1

cancelling $h$      M1

$=3\left({\left(x+h\right)}^2+x\left(x+h\right)+x^2\right)-1$

then $\underset{h\to 0}{\text{lim}}\left(3\left({\left(x+h\right)}^2+x\left(x+h\right)+x^2\right)-1\right)$

$=9x^2-1$      A1

Note: Final A1 dependent on all previous marks.

 

[5 marks]

 

[N/A]