Using the definition of a derivative as \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\) , show that the derivative of \(\frac{1}{{2x + 1}}{\text{ is }}\frac{{ - 2}}{{{{(2x + 1)}^2}}}\).

Prove by induction that the \({n^{{\text{th}}}}\) derivative of \({(2x + 1)^{ - 1}}\) is \({( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\).

let \(f(x) = \frac{1}{{2x + 1}}\) and using the result \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\)

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{1}{{2(x + h) + 1}} - \frac{1}{{2x + 1}}}}{h}} \right)\)     M1A1

\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{[2x + 1] - [2(x + h) + 1]}}{{h[2(x + h) + 1][2x + 1]}}} \right)\)     A1

\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - 2}}{{[2(x + h) + 1][2x + 1]}}} \right)\)     A1

\( \Rightarrow f'(x) = \frac{{ - 2}}{{{{(2x + 1)}^2}}}\)     AG

[4 marks]

let \(y = \frac{1}{{2x + 1}}\)

we want to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = {( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\)

let \(n = 1 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = {( - 1)^1}\frac{{{2^1}1!}}{{{{(2x + 1)}^{1 + 1}}}}\)     M1

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2}}{{{{(2x + 1)}^2}}}\) which is the same result as part (a)

hence the result is true for \(n = 1\)     R1

assume the result is true for \(n = k{\text{ : }}\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = {( - 1)^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}\)     M1

\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}} \right]\)     M1

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}{2^k}k!{{(2x + 1)}^{ - k - 1}}} \right]\)     (A1)

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^k}{2^k}k!( - k - 1){(2x + 1)^{ - k - 2}} \times 2\)     A1

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}{2^{k + 1}}(k + 1)!{(2x + 1)^{ - k - 2}}\)     (A1)

\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}\frac{{{2^{k + 1}}(k + 1)!}}{{{{(2x + 1)}^{k + 2}}}}\)     A1

hence if the result is true for \(n = k\) , it is true for \(n = k + 1\)

since the result is true for \(n = 1\) , the result is proved by mathematical induction     R1 

Note: Only award final R1 if all the M marks have been gained.

[9 marks]

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.

Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.