What is the derivative of #tan^7(x^2)#?

1 Answer
Mar 11, 2018

#f'(x) = 14x(tan^6(x^2))(sec^2(x^2))#

Explanation:

Working with trigonometric functions note that #tan^7(x) = (tan(x))^7.# These two expressions are equivalent. Secondly we must note the Chain Rule. #d/dx f(g(x)) = f'(g(x))*g'(x)#

Let us establish that #tan^7(x^2)# #f(x) = x^7# and #g(x) = tan(x^2).# #f(g(x)) = (tan(x^2))^7 = tan^7(x^2)#Therefore we now have a composition of functions. Following the Power rule for #f(x)# , #d/dx f(x) = x^7 = 7x^6# composing the functions again, #f'(g(x)) = 7(tan^6(x^2)) = 7((tan(x^2))^6)# . We are not finished though.

Now we must find the derivative of #tan(x^2)# . Noting that #tan(x^2)# looks vaguely familiar to another composition of functions. We establish that #f(x) = tan(x)# #g(x) = x^2# . This is the opposite expression/form that our first composition of function occupied. Plugging #f(g(x)) = tan(x^2)# . Finding the derivative of #d/dx tan(x) = sec^2(x)# , so we now have #f'(g(x)) = sec^2(x^2)# , but now we find #d/dx x^2 = 2x# , so in this case from our Chain Rule, #f'(g(x))*g'(x) = 2xsec^2(x^2)# We note that this Composition was in fact #g'(x)# for our original function #tan^7(x^2)#

From this point plug #g'(x)# into our equations giving us #f'(g(x)) *g'(x) = 14x(tan^6(x^2))(sec^2(x^2))# where #7# multiples with #2x = 14x# where #f'(x) = 14x(tan^6(x^2))(sec^2(x^2))#