5.1 Introduction to Differentiation

Topic	5.1 Introduction to Differentiation
Tag	Differentiation, Limits, First principles
Source	M15-TZ1-P1-4(HL)
Question Text	Find the derivative of $f(x)=2 x^3$ using first principles (a) $f^{\prime}(x)=6 x^3$ (b) $f^{\prime}(x)=3 x^2$ (c) $f^{\prime}(x)=6 x^2$ (d) $f^{\prime}(x)=2 x^2$
Total Mark	3
Correct Answer	(c)
Explanation	n/a
Mark Scheme	$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{2(x+h)^3-2 x^3}{h} \\ \lim _{h \rightarrow 0} 2 \frac{x^3+3 x^2 h+3 x h^2+h^3-x^3}{h}=\lim _{h \rightarrow 0} 2 \frac{3 x^2 h+3 x h^2+h^3}{h} \\ \lim _{h \rightarrow 0} 2\left(3 x^2+3 x h+h^2\right)=6 x^2$

Topic	5.1 Introduction to Differentiation
Tag	Differentiation, Limits, First principles
Source	N18/5/MATHL/HP2/ENG/TZ0/XX/5
Question Text	Differentiate from first principles the function $f(x)=4 x^3-2 x$ . (a) $f^{\prime}(x)=12 x^2-2$ (b) $f^{\prime}(x)=x^4-x^2$ (c) $f^{\prime}(x)=12 x^2$ (d) $f^{\prime}(x)=12 x^2-2 x^2$
Total Mark	5
Correct Answer	(a)
Explanation	n/a
Mark Scheme	$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ =\lim _{h \rightarrow 0} \frac{\left(4(x+h)^3-2(x+h)\right)-\left(4 x^3-2 x\right)}{h} \\ =\lim _{h \rightarrow 0} \frac{4\left(x^3+3 x^2 h+3 x h^2+h^3\right)-2 x-2 h-4 x^3+2 x}{h}$ $=\lim _{h \rightarrow 0} \frac{12 x^2 h+12 x h^2+4 h^3-2 h}{h}$ If we cancel $h$ , $\lim _{h \rightarrow 0} 12 x^2+12 x h+4 h^2-2=12 x^2-2$