- #1
subzero0137
- 91
- 4
Work out the first five derivatives of the function [itex]f(x)=sec(x)[/itex], and hence deduce the Maclaurin series of [itex]g(x)=sec(x)(1+tan(x))[/itex] up to and including the term of order [itex]x^4[/itex].
(Hint: why have you been asked for five derivatives of [itex]f(x)[/itex]?)
The Maclaurin series for function [itex]g(x)[/itex] is given by [tex]g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k[/tex]
I know how to differentiate [itex]g(x)[/itex], and although it would take a long time, I could differentiate [itex]g(x)[/itex] 4 times, evaluate the derivatives at [itex]x=0[/itex] and substitute the values in the series equation above to deduce the Maclaurin series up to the [itex]x^4[/itex] term. But I'm not sure how differentiating [itex]f(x)=sec(x)[/itex] 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.
(Hint: why have you been asked for five derivatives of [itex]f(x)[/itex]?)
The Maclaurin series for function [itex]g(x)[/itex] is given by [tex]g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k[/tex]
I know how to differentiate [itex]g(x)[/itex], and although it would take a long time, I could differentiate [itex]g(x)[/itex] 4 times, evaluate the derivatives at [itex]x=0[/itex] and substitute the values in the series equation above to deduce the Maclaurin series up to the [itex]x^4[/itex] term. But I'm not sure how differentiating [itex]f(x)=sec(x)[/itex] 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.
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