Question #8f8da

Question

926 views around the world

You can reuse this answer
Creative Commons License

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