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I stumbled on this calculus problem here:

Let $f(x) = \ln|\sec x + \tan x|$ and $g(x) = \sec x + \tan x.$ Find the fourth derivative of $g(x)$ taken with respect to $f(x)$

A)$\\$ $f'(x)$

B)$\\$ $f'(x)g(x)$

C)$\\$ $g(x)$

D)$\\$ $g'(x)$

E)$\\$NOTA

The answer to the question was C

I understand how to find a function taken with respect with a derivative. For example, if I was trying to find the first derivative:

$$\frac{dg(x)}{df(x)} = \frac{dg(x)}{dx}*\frac{dx}{df(x)}$$

I would simply use the chain rule.

However, for subsequent derivatives, the two functions tend to get more messy, so I decided to look at the solution provided to see if there was an easier method.

Interestingly enough, the solution is:

The first derivative simplifies as follows:

$$\frac{d(g(x))}{d(f(x))} = \frac{d(g(x))}{dx}*\frac{dx}{d(f(x))} = \frac{\sec^2(x) + \sec(x)\tan(x)}{\sec(x)} = g(x)$$ Thus all higher order derivatives will produce the same result, and the answer is $g(x) =$ C.

Is this conclusion true for any function taken with respect with another function? If so, if anyone could point me to a proof that would be also appreciated.

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Varun Iyer
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$$ f = \ln|g(x)|\implies g(x) = \mathrm{e}^f $$ thus $$ g^{(4)} = \dfrac{d^4}{df^4}g = \dfrac{d^4}{df^4}\mathrm{e}^f = \mathrm{e}^f = g(x) $$

Chinny84
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  • So this conclusion cannot be made for all derivatives taken with respect to another function. This relationship was the reason then? – Varun Iyer Jan 16 '15 at 10:10
  • That is correct. The functional form of the exponential allowed for this. for example if it was $f = 2\ln|g(x)|$ then you would not get the relationship of $f^{(n)} = g(x)$ at all. – Chinny84 Jan 16 '15 at 10:12
  • no problems :). I would also like to say thanks for your contributions/answers to the site also. – Chinny84 Jan 16 '15 at 10:19