I am to find the derivative of f(x) and g(x):

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So far, I know the following:

  • The derivative of tan(x) = sec(x)^2
  • The derivative of sec(x) = sec(x)tan(x)

So, I have tried the following steps to get the derivative of f(x):

1) Convert to sec(x) using the chain rule --> (sec(x^4)*4x^3)^2

2) This equals ((4x^3)(sec(x^4))^2

For g(x), however, I am not quite sure where to begin.

As well, I do not believe that ((4x^3)(sec(x^4))^2 = f'(x).

Can anyone please show me where I went wrong trying to find the derivative of f(x), as well as point me in the right direction in finding g(x)?

  • 1
    $\begingroup$ for the second one. use the chain rule (sec(x^5))^3. It will be first power rule, then for sec(x^5), then for x^5 $\endgroup$ – Kishan Kumar Nov 1 '15 at 1:54

Hint. By the chain rule, you rather have $$ (\tan^2(x^4))'=2\times 4x^3\times \sec^2(x^4)\times\tan(x^4) $$ and $$ (\sec^3(x^5))'=3\times 5x^4\times \sec(x^5)\times\tan (x^5)\times\sec^2(x^5) $$


Use the Chain rule three times:

$\frac{d}{dx}($tan$^2(x^4))=(2)$tan$(x^4)\cdot\sec^2(x^4)\cdot 4x^3=\frac{8x^3 sin(x^4)}{cos^3(x^4)}$. Try for the second one.


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