Problem 42 - Calculus of a Single Variable

Question

Find the derivative of the function. \(f(x)=\operatorname{arcsec} 2 x\)

Step-by-Step Solution

Answer

So, the derivative of the function is \(\frac{1}{x\sqrt{4x^2-1}}\)

Step 1: Identify the Main Function and Inside Function

The main function in this scenario is the inverse secant function and the function inside is \(2x\). Therefore, our function can be represented as \(f(g(x))\), where \(f(u)=\operatorname{arcsec}(u)\) and \(g(x)=2x\).

Step 2: Differentiate the Main Function

The derivative of the inverse secant function \(\operatorname{arcsec}(u)\) is given by \(\frac{1}{|u|\sqrt{u^2-1}}\) for \(|u|>1\). So \(f'(g(x))= \frac{1}{|2x|\sqrt{(2x)^2-1}}=\frac{1}{2|x|\sqrt{4x^2-1}}\)

Step 3: Differentiate the Inside Function and Apply the Chain Rule

The derivative of \(2x\) is \(2\). The chain rule in calculus states that the derivative of a composition of functions is the derivative of the outside function times the derivative of the inside function. So, \((f(g(x)))' = f'(g(x)) . g'(x)\). So, the derivative of \(\operatorname{arcsec}(2x)\) is \(\frac{1}{2|x|\sqrt{4x^2-1}}*2=\frac{1}{|x|\sqrt{4x^2-1}}\)

Step 4: Simplify the Result

Looking at the domain of the function, \(2x>=1\) since \(\operatorname{arcsec}(u)\) is only defined for \(|u|>=1\). Thus we can drop the absolute value sign and the expression simplifies to \(\frac{1}{x\sqrt{4x^2-1}}\)