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Chapter 4: Problem 48

\((f(x))^{n}=f(n x)\) Differentiating it \((f(x))^{n-1} f^{\prime}(x)=f^{\prime}(n x)\) Multiply by \(f(x)\) $\mathrm{f}(\mathrm{nx}) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{nx}) \mathrm{f}(\mathrm{x})$

Short Answer

Expert verified
Answer: The relationship between \(f'(x)\) and \(f'(nx)\) is given by the equation \(f(nx) f'(x) = f'(nx) f(x)\).

Step by step solution

01

Apply the Chain Rule to differentiate \((f(x))^n = f(nx)\)

To find the derivative, we will use the Chain Rule, which states that if \(h(x)=g(f(x))\), then \(h'(x)=g'(f(x)) \cdot f'(x)\). Applying this to our given function, we have, \(h(x) = (f(x))^n\), \(g(x) = x^n\), and \(f(x) = f(x)\). So, \(g'(x) = n x^{n-1}\), and the Chain Rule tells us that \(h'(x) = nx^{n-1} f'(x)\).
02

Substitute back the value of \(h(x)\) and \(g(x)\)

We will replace the value of \(h(x)\) and \(g(x)\) in the equation we obtained in step 1: \(nx^{n-1} f'(x) = f'(nx)\). Now we have: \((f(x))^{n-1} f'(x) = f'(nx)\), which is already given in the problem.
03

Multiply both sides of the equation by \(f(x)\)

Multiply both sides of the equation by \(f(x)\) to obtain the relationship between \(f'(x)\) and \(f'(nx)\). \(f(x)(f(x))^{n-1} f'(x) = f'(nx) f(x)\) This simplifies to: \((f(x))^n f'(x) = f'(nx) f(x)\)
04

Substitute back the original equation \((f(x))^n = f(nx)\)

Recall that \((f(x))^n = f(nx)\). Now, replace \((f(x))^n\) with \(f(nx)\) in the equation obtained in step 3: \(f(nx) f'(x) = f'(nx) f(x)\) This is the required relationship between \(f'(x)\) and \(f'(nx)\).

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Most popular questions from this chapter

\(y=\log _{e^{t}}(x-2)^{2}\) \(y=\frac{2 \log (x-2)}{\log x}\) $y^{\prime}=\frac{2(\log x) \times \frac{1}{x-2}-2 \log (x-2) \times \frac{1}{x}}{(\log x)^{2}}$ \(y^{\prime}=\frac{2}{\log 3}\) at \(x=3\)

If \(\mathrm{f}(\mathrm{x})=|\ln | \mathrm{x} \|\), then \(\mathrm{f}^{\prime}(\mathrm{x})\) equals (A) \(\frac{-\operatorname{sgn} \mathrm{x}}{|\mathrm{x}|}\), for \(|\mathrm{x}|<1\), where \(\mathrm{x} \neq 0\) (B) \(\frac{1}{x}\) for \(|x|>1\) and \(-\frac{1}{x}\) for \(|x|<1, x \neq 0\) (C) \(-\frac{1}{x}\) for \(|x|>1\) and \(\frac{1}{x}\) for \(|x|<1\) (D) \(\frac{1}{x}\) for \(|x|>0\) and \(-\frac{1}{x}\) for \(x<0\)

\(3 \mathrm{f}(\cos x)+2 f(\sin x)=5 x\) Put \(x=\frac{\pi}{2}-x\) \(3 \mathrm{f}(\sin x)+2 f(\cos x)=\frac{5 \pi}{2}-5 x\) Solving (I) \& (II) \(5 f(\cos x)=25 x-5 \pi\) \(f(\cos x)=5 x-\pi\) \(f^{\prime}(\cos x)=\frac{-5}{\sin x}\)

$\lim _{x \rightarrow 1} \frac{n x^{n+1}-(n+1) x^{n}+1}{\left(e^{x}-e\right) \sin \pi x}$ $\lim _{h \rightarrow 0} \frac{n(1+h)^{n+1}-(n+1)(1+h)^{n}+1}{-e\left(e^{h}-1\right) \sin \pi h}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(1+\mathrm{h})^{\mathrm{n}+1}-(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}}+1}{-\pi \mathrm{t} \mathrm{h}^{2}}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}}-\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}-1}}{-2 \pi \mathrm{t} h}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n})(1+\mathrm{h})^{\mathrm{n}-1}-\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}-2}(\mathrm{n}-1)}{-2 \pi \mathrm{e}}$ $=\frac{\mathrm{n}^{2}(\mathrm{n}+1)-\mathrm{n}(\mathrm{n}-1)}{-2 \pi \mathrm{e}}(\mathrm{n}+1)=\frac{\mathrm{n}(\mathrm{n}+1)}{-2 \pi \mathrm{e}}$ Put \(\mathrm{n}=100\) \(=\frac{-5050}{\pi \mathrm{e}}\)

$\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}$ Applying C \& D, $\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\mathrm{f}(2 \mathrm{x}-2 \mathrm{y})}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\sin (\mathrm{x}-\mathrm{y})}$ \(\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\sin (\mathrm{x}+\mathrm{y})}\) is constant \(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{C} \sin \frac{\mathrm{x}}{2}\) \(\mathrm{C}=1\) using \(\mathrm{f}^{\prime}(0)=\frac{1}{2}\) \(\Rightarrow \mathrm{f}(\mathrm{x})=\sin \frac{\mathrm{x}}{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cos \frac{\mathrm{x}}{2}\) $\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-1}{4} \sin \frac{\mathrm{x}}{2}$ $\Rightarrow 4 \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=0$
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