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Chapter 6: Problem 22

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

Short Answer

Expert verified
So, \(\frac{d(f(f(x)))}{d x} = \frac{1}{{(x - 1)}^{2} \cdot {(-x + 2)}^{2}}\).

Step by step solution

01

Find the derivative of the function \(f(x)\)

To do so, use the power rule, which states that the derivative of \(x^n\) is \(n*x^{n-1}\). When differentiating \(f(x)=\frac{1}{x-1}\), it is better to rewrite it as \(f(x) = (x - 1)^{-1}\). Therefore, the derivative would be \(f'(x) = -1 * (x - 1)^{-2}\), which simplifies to \(f'(x) = -\frac{1}{{(x - 1)}^{2}}\).
02

Find the derivative of the composed function

To find \(\frac{d(f(f(x)))}{d x}\), we can use the chain rule. The chain rule states that the derivative of a composed function \(f(g(x))\) is \(f'(g(x)) * g'(x)\), i.e., the derivative of the outer function evaluated at the function inside, multiplied by the derivative of the inner function. In our case the 'inner' and the 'outer' function are both \(f\). Therefore we have \(\frac{d(f(f(x)))}{d x} = f'(f(x))\cdot f'(x) = -\frac{1}{{(f(x) - 1)}^{2}} \cdot -\frac{1}{{(x - 1)}^{2}} = \frac{1}{{(x - 1)}^{2} \cdot {(f(x) - 1)}^{2}}\).
03

Substitute the original function for \(f(x)\) in Step 2

Substituting \(f(x)=\frac{1}{x-1}\) in the result of Step 2, the final answer becomes \(\frac{1}{{(x - 1)}^{2} \cdot {(\frac{1} {x - 1} - 1)}^{2}} = \frac{1}{{(x - 1)^2} * {(1- x + 1)}^{2}} = \frac{1}{{(x - 1)}^{2} \cdot {(-x + 2)}^{2}}\).

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Most popular questions from this chapter

\(f(x)=x^{\alpha} \sin \left(\frac{1}{x}\right), x \neq 0, f(0)=0\) 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) \(5 / 3\)

The number of roots of the equation \(\sin x+2 \sin 2 x\) \(+3 \sin 3 x=0\) is the interval \([0, \pi]\) is (a) 1 (b) 2 (c) 3 (d) More than 3

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If \(y=\tan ^{-1}\left\\{\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right\\}\), find \(\frac{d y}{d x}\).
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