Finding derivatives of inverse trig functions using logarithms

[MadAtom]

For some polynomial functions it is useful to logarithmize both sides of the eq. First. How can this be applied for inverse trig functions? Is it even possible?

 

[Mark44]

For some polynomial functions it is useful to logarithmize both sides of the eq. First. How can this be applied for inverse trig functions? Is it even possible?

I can't see how this would be useful for inverse trig functions, or even how it would be useful for polynomials. Polynomials generally consist of a sum of terms, and there is no property that let's you simplify the log of a sum.

 

[MadAtom]

Sorry for that. I meant product and quotient of polynomial expressions.

Now About the inverse trig function: I saw in a book a list of exercises where they said to apply that logarithmization technique and there were some inverse trig functions, but I it was a mistake: They probably mixed topics...

 

[haruspex]

I can't see how this would be useful for inverse trig functions, or even how it would be useful for polynomials. Polynomials generally consist of a sum of terms, and there is no property that let's you simplify the log of a sum.

It is useful if the polynomial is in the form of a product linear expressions.

 

[SammyS]

Do you suppose they were referring to the multiplicative inverse? ... such as getting the derivative of secant from cosine, or cotangent from tangent, etc.

If y = sec(x), then

ln(y) = -ln(cos(x))

Differentiating gives:

$\displaystyle \frac{y'}{y}=-\frac{-\sin(x)}{\cos(x)}$

...​

It seems not much an improvement over just treating sec(x) as 1/cos(x) & using the chain rule .