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))