Question #8f8da

Question

1136 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-18T01:48:02

$π \cdot {sec}^{2} (π \cdot x)$ $pi*sec^2(pi*x)$

$tan (π \cdot x) = \frac{sin (π \cdot x)}{cos (π \cdot x)}$ $tan(pi*x) = sin(pi*x)/cos(pi*x)$

...so, you can use the rule for finding the derivative of the quotient of two functions.

if $f (x) = \frac{u (x)}{v (x)}$ $f(x) = (u(x))/(v(x))$ , then $f'(x) = (u'(x)v(x)-u(x)v'(x))/(v(x)^2)$

so, for this function, you'd have:

$(pi*cos(pi*x)cos(pi*x) + sin(pi*x)pi sin(pi*x))/cos^2(p*x)$

Which simplifies to:

$(pi(cos^2(pi*x) + sin^2(pi*x)))/cos^2(pi*x)$

since $cos^2(a) + sin^2(a) = 1$ , we can further simplify to:

$pi/cos^2(pi*x)$

and that is:

$pi * sec^2(pi*x)$

GOOD LUCK