How do you find the derivative of #y=tan^2(5x)# ?

1 Answer

The derivative of tan2(5x) is 10tan(5x)sec2(5x).

Explanation:

The derivative of #tan^2(5x)# is #10tan(5x)sec^2(5x)#.

Our function is the composite of three simpler functions:
Start with #x# and multipy it by #5#.
Then find the tangent of that.
Finally, find the square of that.

So to find the derivative we need the threefold chain rule:

If #k(x) = f(g(h(x)))#, then
#k'(x) = f'((g(h(x))*g'(h(x))*h'(x)#.

In our problem, #f(x) = x^2, g(x) = tan(x)#, and #h(x) = 5x#,
so #f'(x) = 2x, g'(x) = sec^2(x)# and #h'(x) = 5#

Thus #k'(x) = f'((g(h(x))*g'(h(x))*h'(x) #
#= f'(tan5x)*g'(5x)*h'(x)#
#=2tan(5x)*sec^2(5x)*5#
#=10tan(5x)sec^2(5x)#

Note: You might prefer doing the problem by letting
#u=5x, v=tan u, y = v^2#

Then
#dy/dx = ((dy)/(dv))((dv)/(du))((du)/(dx))#
#=2v*sec^2u*5#
#=10tan(5x)sec^2(5x)#

Many students prefer the second method, but occasionally you may encounter a problem for which the notation in the second method becomes ambiguous, so it is good to know both methods.