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Chapter 15: Problem 6

Find the third and fourth derivatives of \(y\) given the second derivative of \(y\) is (a) \(\frac{2}{\mathrm{e}^{3 x}}\) (b) \(\frac{1+x}{x^{2}}\) (c) \(3 \ln x^{2}\) (d) \(\sin x+\sin (-2 x)\) (e) \(\frac{\cos ^{2} x+\cos x}{\cos x}\)

Short Answer

Expert verified
To summarize, the third and fourth derivatives for each of the given functions are: (a) \(y'''=-6\mathrm{e}^{-3x}\) and \(y^{(4)}=18\mathrm{e}^{-3x}\). (b) \(y'''=\frac{2x^{2}-3x-1}{x^{4}}\) and \(y^{(4)}=\frac{4x^{3}-32x^{2}+24x+4}{x^{8}}\). (c) \(y'''=6x^{-1}\) and \(y^{(4)}=-6x^{-2}\). (d) \(y'''=\cos x -2\cos(-2x)\) and \(y^{(4)}=-\sin x +4\sin(-2x)\). (e) \(y'''=\frac{-3\cos x+\sin x+1}{\cos^{2} x}\) and \(y^{(4)}=\frac{2\sin x(\cos x-\sin ^{2} x+2)}{(1-\cos^{2} x)^2}\).

Step by step solution

01

Find the third derivative (

Differentiate the given function one more time.)
02

Find the fourth derivative (

Differentiate the third derivative.) ## (a) \(y''=\frac{2}{\mathrm{e}^{3 x}}\)
03

Find the third derivative (

Differentiate with respect to x, \(y'''=\frac{d(\frac{2}{\mathrm{e}^{3 x}})}{dx}\).) Using the chain rule, where \(u=\mathrm{e}^{3x}\), we have: \(y'''=\frac{d(\frac{2}{u})}{dx}=\frac{-2\frac{du}{dx}}{u^{2}}=\frac{-6\mathrm{e}^{3x}}{(\mathrm{e}^{3x})^2}=-6\mathrm{e}^{-3x}.\)
04

Find the fourth derivative (

Differentiate with respect to x, \(y^{(4)}=-6\frac{d(\mathrm{e}^{-3x})}{dx}\).) Using the chain rule, where \(v=-3x\), we have: \(y^{(4)}=-6\frac{d(\mathrm{e}^{v})}{dx}=-6\mathrm{e}^{v}\frac{dv}{dx}=18\mathrm{e}^{-3x}.\) ## (b) \(y''=\frac{1+x}{x^{2}}\)
05

Find the third derivative (

Differentiate with respect to x, \(y'''=\frac{d(\frac{1+x}{x^{2}})}{dx}\).) Applying the quotient rule, where \(p=x+1\) and \(q=x^{2}\), we have: \(y'''=\frac{p'(x)q(x)-p(x)q'(x)} {(q(x))^{2}}=\frac{2x^{2}-3x-1}{x^{4}}\).
06

Find the fourth derivative (

Differentiate with respect to x, \(y^{(4)}=\frac{d(\frac{2x^{2}-3x-1}{x^{4}})}{dx}\).) Applying the quotient rule, where \(r=2x^{2}-3x-1\) and \(s=x^{4}\), we have: \(y^{(4)}=\frac{r'(x)s(x)-r(x)s'(x)} {(s(x))^{2}}=\frac{4x^{3}-32x^{2}+24x+4}{x^{8}}\). ## (c) \(y''=3 \ln x^{2}\)
07

Find the third derivative (

Differentiate with respect to x, \(y'''=\frac{d(3 \ln x^2)}{dx}\).) Using the chain rule, where \(w=x^2\), we have: \(y'''=3\frac{d(\ln w)}{dx}=3\frac{1}{w}\frac{dw}{dx}=6x^{-1}\).
08

Find the fourth derivative (

Differentiate with respect to x, \(y^{(4)}=\frac{d(6x^{-1})}{dx}\).) Thus, \(y^{(4)}=-6x^{-2}\). ## (d) \(y''=\sin x+\sin(-2x)\)
09

Find the third derivative (

Differentiate with respect to x, \(y'''=\frac{d(\sin x + \sin(-2x))}{dx}\).) Differentiating the sum of the two functions, we have: \(y'''=\cos x -2\cos(-2x)\).
10

Find the fourth derivative (

Differentiate with respect to x, \(y^{(4)}=\frac{d(\cos x -2\cos(-2x))}{dx}\).) Differentiating the difference of the two functions, we have: \(y^{(4)}=-\sin x +4\sin(-2x)\). ## (e) \(y''=\frac{\cos^{2} x+\cos x}{\cos x}\)
11

Find the third derivative (

Differentiate with respect to x, \(y'''=\frac{d(\frac{\cos^{2} x+\cos x}{\cos x})}{dx}\).) Applying the quotient rule, where \(m=\cos^{2} x+\cos x\) and \(n=\cos x\), we have: \(y'''=\frac{m'(x)n(x)-m(x)n'(x)} {(n(x))^{2}}=\frac{-3\cos x+\sin x+1}{\cos^{2} x}\).
12

Find the fourth derivative (

Differentiate with respect to x, \(y^{(4)}=\frac{d(\frac{-3\cos x+\sin x+1}{\cos^{2} x})}{dx}\).) Applying the quotient rule, where \(f=-3\cos x+\sin x+1\) and \(g=\cos^{2} x\), we have: \(y^{(4)}=\frac{f'(x)g(x)-f(x)g'(x)} {(g(x))^{2}}=\frac{2\sin x(\cos x-\sin ^{2} x+2)}{(1-\cos^{2} x)^2}\). And that's it! We have found the third and fourth derivatives for each of the given functions.

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Find the rate of change of the following functions: (a) \(\frac{3 t^{3}-t^{2}}{2 t}\) (b) \(\ln \sqrt{x}\) (c) \((t+2)(2 t-1)\) (d) \(\mathrm{e}^{3 v}\left(1-\mathrm{e}^{v}\right)\) (e) \(\sqrt{x}(\sqrt{x}-1)\)

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