# Maclaurin series

Work out the first five derivatives of the function $f(x)=sec(x)$, and hence deduce the Maclaurin series of $g(x)=sec(x)(1+tan(x))$ up to and including the term of order $x^4$.

(Hint: why have you been asked for five derivatives of $f(x)$?)

The Maclaurin series for function $g(x)$ is given by $$g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k$$

I know how to differentiate $g(x)$, and although it would take a long time, I could differentiate $g(x)$ 4 times, evaluate the derivatives at $x=0$ and substitute the values in the series equation above to deduce the Maclaurin series up to the $x^4$ term. But I'm not sure how differentiating $f(x)=sec(x)$ 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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## Answers and Replies

Mark44
Mentor
Work out the first five derivatives of the function $f(x)=sec(x)$, and hence deduce the Maclaurin series of $g(x)=sec(x)(1+tan(x))$ up to and including the term of order $x^4$.

(Hint: why have you been asked for five derivatives of $f(x)$?)

The Maclaurin series for function $g(x)$ is given by $$g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k$$

I know how to differentiate $g(x)$, and although it would take a long time, I could differentiate $g(x)$ 4 times, evaluate the derivatives at $x=0$ and substitute the values in the series equation above to deduce the Maclaurin series up to the $x^4$ term. But I'm not sure how differentiating $f(x)=sec(x)$ 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.

This is a Maclaurin series, so 1) each derivative is evaluated at x = 0, and 2) it's a sum of powers of x. Once you get to the 5th derivative of sec(x) I think you'll see what the hint is about.

This is a Maclaurin series, so 1) each derivative is evaluated at x = 0, and 2) it's a sum of powers of x. Once you get to the 5th derivative of sec(x) I think you'll see what the hint is about.

The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.

Mark44
Mentor
The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.
That looks fine. So with f(x) = sec(x), you have f(5)(x) = tan(x)sec(x)(120sec^4(x)-60sec^(x)+1).
What is f(5)(0)?

Dick
Homework Helper
The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.

I think the sense of the hint is that you should compute the Maclaurin series of sec(x) up to power x^4 and then multiply by the Maclaurin series of 1+tan(x), rather than computing the series of sec(x)(1+tan(x)) directly. The differentiation is easier.

That looks fine. So with f(x) = sec(x), you have f(5)(x) = tan(x)sec(x)(120sec^4(x)-60sec^(x)+1).
What is f(5)(0)?

f(5)(0) = 0

I think the sense of the hint is that you should compute the Maclaurin series of sec(x) up to power x^4 and then multiply by the Maclaurin series of 1+tan(x), rather than computing the series of sec(x)(1+tan(x)) directly. The differentiation is easier.

Oh, I see. But to compute the Maclaurin series for sec(x) up to the power x^4, I only need to differentiate it 4 times. Why did the question ask for the 5th derivative of sec(x)?

vela
Staff Emeritus
Homework Helper
I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.

I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.

Ohhhh! How did I miss that... :surprised

I now understand why they asked for 5 derivatives of sec(x). It's so that I can obtain the x4 order term for the Maclaurin series of ##f'(x)##.

Thanks for the replies, everyone :)

Dick