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

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Psykolord1989 . · Amory W.
Aug 29, 2014

The derivative of #y=tan(x^2)# is #(dy)/(dx) = (2x)[sec^2(x^2)]#

In solving this problem, we will have to make use of the chain rule, which states that for a function #y = f(g(x))# (that is, #y# is a function which is itself a function of another function), #y'(x) = (g'(x))*f'(g(x))#.

In this case, #f(g(x)) = tan(x^2)#. Our #g(x) = x^2# and our #f(g) = tan(g(x))#. Thus, if our function #f(g)# were #f(x)# we would have #f(x) = tan(x)#.

Via the power rule we know that #d/(dx) (x^2) = 2x#, and via our definitions of the derivatives of trigonometric functions, we know that #f'(x) = sec^2(x)#,or #f'(g) = sec^2(g)#

Thus, by using the chain rule shown above, we obtain:

#y'(x) = (2x)sec^2(x^2)#

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