# Thread: finding the second derivitive

1. ## finding the second derivitive

I had some problems trying to find the second derivative of some equations.
1. sec(x). I used the product rule when i got to the first deriv. but my answer came as sec x^3 which is incorrect and it says: sec(x)^3+secx*tanx^2 as the right answer.
2. e^x*cos x. Again I used the product rule and i got -e^x*cos x+e^x*cos x which is wrong and it says -2e^x*sin x is the right answer.
3.(sqrt x) I had a problem with the distribution and exponents. Anyways, I got x^1/2*(1/x)+(1/2 x^-1/2)*ln x. the right answer is -1/4*x^-3/2 *ln t.

2. Originally Posted by driver327
1. sec(x). I used the product rule when
Product rule?

This is a basic derivative, which is $\displaystyle (\sec{x})'=\sec{x}\tan{x}$, now to find the second derivative just use carefully the product rule.

3. $\displaystyle 1.)(\sec(x))''$
$\displaystyle = (\sec(x)\tan(x))'$
$\displaystyle = \sec(x)'\tan(x) + '\sec(x)\tan(x)$
$\displaystyle = \sec(x)\sec^2(x) + \sec(x)\tan(x)\tan(x)$
$\displaystyle = \sec^3{x} + \sec{x}\tan^2{x}$

$\displaystyle 2.)(e^{x}\cos(x))''$
$\displaystyle = (-e^{x}sin{x} + e^{x}cos{x})'$
$\displaystyle = -e^{x}\sin{x} -e^{x}\cos + e^x\cos{x} - e^{x}\sin{x}$
cancel the cosine
$\displaystyle = -2e^{x}\sin{x}$

4. $\displaystyle 3.) (\sqrt{x})''$
$\displaystyle =(\frac{1}{2\sqrt{x}})'$
$\displaystyle =\frac{1}{2}(x^{-\frac{1}{2}})$
$\displaystyle =\frac{1}{2}(-\frac{1}{2}(x^{-\frac{1}{2}-1}))$
$\displaystyle =-\frac{1}{4}(x^{-\frac{3}{2}})$
$\displaystyle =-\frac{1}{4x^{\frac{3}{2}}}$