1. ## differention 2

differentiate this:

$\displaystyle f (x) = xsecx$

$\displaystyle f '(x) = (1)(secx) + (x)(tanxsecx) = secx + xtanxsecx$

2. Originally Posted by jvignacio
differentiate this:

$\displaystyle f (x) = xsecx$

$\displaystyle f '(x) = (1)(secx) + (x)(tanxsecx) = secx + xtanxsecx$

Yes, you are all right using the product rule.

3. Originally Posted by Shyam
Yes, you are all right using the product rule.

f '(x) = secx + xtanxsecx

4. Originally Posted by jvignacio

f '(x) = secx + xtanxsecx
Yes, it is correct.

It looks neater written as $\displaystyle f'(x) = \sec x \left(1 + x\tan x \right)$.

5. Originally Posted by Air
Yes, it is correct.

It looks neater written as $\displaystyle f'(x) = \sec x \left(1 + x\tan x \right)$.
how did u get it in that form?

6. Originally Posted by jvignacio
how did u get it in that form?
You have:

$\displaystyle f'(x) = \sec x + x \sec x \tan x$

There is a common $\displaystyle \sec x$ in the equation so factorise it out.

$\displaystyle f'(x) = \sec{x} \left( 1 + x\tan x \right)$

You still get full marks for writing it in your form but it looks neater to have it factorised.

7. Originally Posted by Air
You have:

$\displaystyle f'(x) = \sec x + x \sec x \tan x$

There is a common $\displaystyle \sec x$ in the equation so factorise it out.

$\displaystyle f'(x) = \sec{x} \left( 1 + x\tan x \right)$

You still get full marks for writing it in your form but it looks neater to have it factorised.

thank you!!!