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Answer 1 · 2017-10-18T01:48:02

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

$tan(pi*x) = sin(pi*x)/cos(pi*x)$

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

if $f(x) = (u(x))/(v(x))$ , 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