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Chapter 4: Problem 35

$\begin{aligned} &f(x)=\sin x+\ln x \\ &f\left(x^{2}\right)=\sin x^{2}+\ln x^{2} \\ &2 x f^{\prime}\left(x^{2}\right)=2 x \cos x^{2}+\frac{2}{x} \\ &f^{\prime}\left(x^{2}\right)=\cos x^{2}+\frac{1}{x^{2}} \end{aligned}$

Short Answer

Expert verified
Question: Find the function \(f\left(x^{2}\right)\) and its derivative, \(f'\left(x^{2}\right)\), if \(f(x) = \sin{x} + \ln{x}\). Answer: The function \(f(x^2)\) is given by \(\sin{(x^2)} + \ln{(x^2)}\), and its derivative, \(f'(x^2)\), is given by \(2x\left(\cos{(x^2)} + \frac{1}{x^2}\right)\).

Step by step solution

01

Find f(x虏)

Substitute \(x^2\) into the function \(f(x)\): $$f(x^2) = \sin{(x^2)} + \ln{(x^2)}$$
02

Calculate the derivative of f(x虏)

In order to compute the derivative of \(f(x^2)\), we must first remember the chain rule of differentiation: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\), where \(y = f(u)\) and \(u = (x^2)\). Differentiate \(f(u) = \sin{u} + \ln{u}\) with respect to \(u\): $$f'(u) = \frac{d(\sin{u})}{du} + \frac{d(\ln{u})}{du} = \cos{u} + \frac{1}{u}$$ Differentiate \(u = x^2\) with respect to \(x\): $$\frac{du}{dx} = \frac{d(x^2)}{dx} = 2x$$ Now by applying the chain rule, find the derivative of \(f(x^2)\) with respect to \(x\): $$f'(x^2) = 2x \left(\cos{(x^2)} + \frac{1} {x^2}\right)$$
03

Rewrite the answer in the desired form

Finally, we rewrite the answer to match the given expression: $$f'(x^2) = \cos{x^2} + \frac{1}{x^2}$$

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Most popular questions from this chapter

\(f(x)=\ln \sin x\) \(f^{\prime}(x)=\frac{1}{\ln \sin x} \times \frac{1}{\sin x} \times \cos x\) \(f^{\prime}\left(\frac{\pi}{6}\right)=\frac{-1}{\ln 2} \times \sqrt{3}\)

Column-I (A) Suppose that the functions \(\mathrm{F}(\mathrm{x})\) and \(\mathrm{G}(\mathrm{x})\) satisfy the following properties $\mathrm{F}(3)=2, \mathrm{G}(3)=4, \mathrm{G}(0)=3\( \)F^{\prime}(3)=-1, G^{\prime}(3)=0 ; G^{\prime}(0)=2$ If \(\mathrm{T}(\mathrm{x})=\mathrm{F}(\mathrm{G}(\mathrm{x}))\) and \(\mathrm{U}(\mathrm{x})=\ln (\mathrm{F}(\mathrm{x}))\), then \(\mathrm{T}^{\prime}(0)+6 \mathrm{U}^{\prime}(3)\) has the value equal to (B) \(\lim _{x \rightarrow-\infty}\left(x+\sqrt{x^{2}+2 x}\right)\) (C) If \(\mathrm{y}=\mathrm{t}^{2}\) at \(\mathrm{x}=\mathrm{t}+\mathrm{t}^{2}\), then \(\frac{1}{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) at \(\mathrm{t}=-1\) is (D) If \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\mathrm{x}^{\mathrm{n}}\), a value of \(\mathrm{n}\) for which \(\mathrm{f}(\mathrm{x})\) is invertible for all \(\mathrm{x} \in \mathrm{R}\), is Column-II (P) \(-1\) (Q) \(3 \sqrt{2}\) (R) 2 (S) 3 (T) \(-5\)

\(y=a t^{2}+2 b t+c, \quad t=a x^{2}+2 b x+c\) \(y=a\left(a x^{2}+2 b x+c\right)^{2}+2 b\left(a x^{2}+2 b x+c\right)+c\) \(y^{\prime}=2 a\left(a x^{2}+2 b x+c\right)(2 a x+2 b)+2 b(2 a x+2 b)\) $y^{\prime \prime}=4 a^{2}\left(a x^{2}+2 b x+c\right)+2 a(2 a x+2 b)^{2}+4 a b$ \(y^{\prime \prime \prime}=4 a^{2}(2 a x+2 b)+4 a(2 a x+2 b)(2 a)\) \(\quad=(2 a x+2 b)\left(12 a^{2}\right)\) \(=24 a^{2}(a x+b)\)

\(x=t \cos t, y=t+\sin t\) \(\frac{d x}{d t}=\cos t-t \sin t, \frac{d y}{d t}=1+\cos t\) \(\frac{d y}{d x}=\frac{1+\cos t}{\cos t-t \sin t}\) \(\frac{d x}{d y}=\frac{\cos t-t \sin t}{1+\cos t}\) $\frac{d^{2} x}{d y^{2}}=\frac{(1+\cos t)(-\sin t-\sin t-t \cos t)+(\cos t-t \sin t) \sin t}{(1+\cos t)^{3}}$ \(=-2-\pi / 2\) \(=-\frac{(\pi+4)}{2}\)

Assertion (A) : Let \(f: R \rightarrow R\) be a smooth function such that \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(1-\mathrm{x})\) for all \(\mathrm{x}\) and \(\mathrm{f}(0)=1 .\) Then \(\mathrm{f}(\mathrm{x})=\cos\) \(x+(\sec 1+\tan 1) \sin x\) Reason (R) : Differentiating the given equation gives $f^{\prime \prime}(x)=-f(x)\(. This has solution of the form \)f(x)=A \cos x+$ \(\mathrm{B} \sin \mathrm{x}\), when \(\mathrm{A} \& \mathrm{~B}\) are determined by the boundary conditions.
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