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Chapter 6: Problem 23

If \(f(x)=\frac{x-1}{x+1}\), find \(\frac{d(f(f(f(x))))}{d x}\)

Short Answer

Expert verified
The derivative of \(f(f(f(x)))\) with respect to \(x\) is \[\frac{8}{(f(f(x))+1)^2 \cdot (f(x)+1)^2 \cdot (x+1)^2}\]

Step by step solution

01

Understand the function

Here the function given is \(f(x) = \frac{x-1}{x+1}\). First, we need to understand the function and its general behaviour before proceeding to finding its derivative.
02

Find the derivative of the function

The derivative of \(f(x)\) can be calculated using the quotient rule which says that the derivative of \(\frac{u}{v} = \frac{v \cdot (du/dx) - u \cdot (dv/dx)}{v^2}\) where \(u\) and \(v\) are functions of \(x\). Here, \(u = x-1\) and \(v = x+1\). So, \(du/dx = 1\) and \(dv/dx = 1\).\nThus, \[f'(x) = \frac{(x+1)\cdot 1 - (x-1)\cdot 1}{(x+1)^2} = \frac{2}{(x+1)^2}\]
03

Apply the chain rule

The chain rule says that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Here we want to find \(\frac{d(f(f(f(x))))}{d x} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)\).\nLet's first find f'(f(x)). As we have found that \(f'(x) = \frac{2}{(x+1)^2}\), we find that \(f'(f(x)) = \frac{2}{(f(x)+1)^2}\).\nSimilarly, \(f'(f(f(x))) = \frac{2}{(f(f(x))+1)^2}\).\nSubstituting these values: \[\frac{d(f(f(f(x))))}{d x} = \frac{2}{(f(f(x))+1)^2} * \frac{2}{(f(x)+1)^2} * \frac{2}{(x+1)^2}\]

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Most popular questions from this chapter

If \(x^{y}=e^{x-y}\), prove that, \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}} .\)

\(f(x)=x^{\alpha} \sin \left(\frac{1}{x}\right), x \neq 0, f(0)=0\) satisfies conditions of Rolle's theorem on \(\left[-\frac{1}{\pi}, \frac{1}{\pi}\right]\) for \(\alpha\) equals (a) \(-1\) (b) 0 (c) \(7 / 2\) (d) \(5 / 3\)

Find \(\frac{d y}{d x}\), when \(x=a(t-\sin t), y=a(l-\cos t) .\)

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