# Chain rule for trig function

• May 5th 2009, 07:21 AM
tom ato
Chain rule for trig function
Find the second derivative:

$
f(x) = \frac{1}{x} + \tan(x)
$

$
f'(x) = -x^{-2} + \sec^2(x)
$

$
f"(x) = 2x^{-3} + 2\sec(x)\sec(x)\tan(x)
$

$
f"(x) = \frac{2}{x^3} + 2\sec^2(x)\tan(x)
$

My teacher got $f"(x) = \frac{2}{x^3} + 2\sec^2(x)\tan^2(x)$ but I think she just put the tanx in for u on accident... Is my second derivative correct?
• May 5th 2009, 07:28 AM
Quote:

Originally Posted by tom ato
Find the second derivative:

$
f(x) = \frac{1}{x} + \tan(x)
$

$
f'(x) = -x^{-2} + \sec^2(x)
$

$
f"(x) = 2x^{-3} + 2\sec(x)\sec(x)\tan(x)
$

$
f"(x) = \frac{2}{x^3} + 2\sec^2(x)\tan(x)
$

My teacher got $f"(x) = \frac{2}{x^3} + 2\sec^2(x)\tan^2(x)$ but I think she just put the tanx in for u on accident... Is my second derivative correct?

I think you are correct.