1) Obtain the first two terms of the Maclaurin series for...

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00:01 Hi, from the question given that f of x is equal to e to the power of 3x sin x.

00:13 So, here we need to find the maclaurin series for the first non -zero 4 terms.

00:18 So, in general we know that maclaurin series can be written as f of x is equal to f of 0 plus f dash of 0 divided by 1 factorial x plus f double dash of 0 divided by 2 factorial x squared plus f triple dash of 0 divided by 3 factorial x cube plus and so on.

00:46 So, here f of x is equal to e to the power of 3x sin x.

00:54 So, this implies f of 0 is equal to sin 0 is 0, so 0.

00:59 Now, f dash of x is equal to by using the product rule.

01:04 So, e to the power of 3x differentiation of sin x is cos x plus sin x e to the power of differentiation of e to the power of 3x is 3 times of e to the power of 3x.

01:19 Now, f dash of 0 is equal to here sin 0 is 0.

01:24 So, here we have e to the power of 0 is 1 cos 0 is 1.

01:28 So, 1.

01:29 Now, f double dash of x.

01:33 So, again use the product rule for the first and the second term.

01:37 So, we have e to the power of 3x differentiation of cos x is minus sin x plus cos x 3 e to the power of 3x plus 3 e to the power of 3x differentiation of sin x is cos x plus sin x 9 e to the power of 3x.

02:07 Now, f double dash of 0.

02:11 So, here we have sin term and here is also sin term...

Question

1) Obtain the first two terms of the Maclaurin series for sin(tan x). A) $x - \frac{x^3}{3}$ B) $x - \frac{x^3}{6}$ C) $x + \frac{x^3}{3}$ D) $x + \frac{x^3}{6}$

Question

1) Obtain the first two terms of the Maclaurin series for sin(tan x). A) $x - \frac{x^3}{3}$ B) $x - \frac{x^3}{6}$ C) $x + \frac{x^3}{3}$ D) $x + \frac{x^3}{6}$

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