Ok, this is a relatively simple application of the product rule if we know the derivatives of tan(x) and sec(x) but we'll derive them to give it a bit of challenge.
#color(blue)("Optional Derivation Interlude")#
Remember that #tan(x) = (sin(x))/(cos(x))#
#d/(dx)(tan(x)) = d/(dx)((sin(x))/(cos(x)))#
Going to use the quotient rule here:
#d/(dx)((f(x))/(g(x))) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2#
#d/(dx)(tan(x)) = (cos(x)cos(x) - sin(x)(-sin(x)))/(cos(x))^2 = (cos^2(x) + sin^2(x))/(cos^2(x))#
Recall that #sin^2x + cos^2x = 1# so:
#d/(dx)(tan(x)) = 1/(cos^2(x)) = sec^2(x)#
#d/(dx)(sec(x)) = d/(dx)(1/(cos(x)))#
Quotient rule time again:
#(0*cos(x) - 1*(-sin(x)))/(cos^2(x)) = (sin(x))/(cos^2(x)) = (sin(x))/(cos(x))(1)/(cos(x))#
#therefore d/(dx)(sec(x)) = tan(x)sec(x)#
#color(blue)("Derivation End")#
Product rule given by:
#d/(dx)(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)#
#d/(dx)(tan(x)sec(x)) = d/(dx)(tan(x))sec(x) + tan(x)d/(dx)(sec(x))#
#=sec^2(x)sec(x) + tan(x)tan(x)sec(x)#
#=sec^3(x) + tan^2(x)sec(x) = sec(x)(sec^2(x)+tan^2(x))#
