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Question Number 209023 by Spillover last updated on 30/Jun/24

Commented by Spillover last updated on 01/Jul/24

$$letu=\frac{4x}{1+5x}\frac{du}{dx}=\frac{4}{{(1+5x)}^2}$$$$\frac{d}{dx}\left({\tan }^{-1}\frac{4x}{1+5x}\right)=\frac{\frac{4}{{(1+5x)}^2}}{1+{\left(\frac{4x}{1+5x}\right)}^2}=\frac{4}{1+10x+25x^2}$$$$also$$$$\frac{d}{dx}\left({\tan }^{-1}\frac{2+3x}{3-2x}\right)u=\frac{2+3x}{3-2x}useqountientrule$$$$\frac{du}{dx}=\frac{13+6x}{{(3-2x)}^2}$$$$\frac{d}{dx}\left({\tan }^{-1}\frac{2+3x}{3-2x}\right)=\frac{13+6x}{13+13x^2}=\frac{13+6x}{13(1+x^2)}$$$$then$$$$=\frac{4}{1+10x+25x^2}+\frac{13+6x}{13(1+x^2)}$$$$=\frac{5}{1+25x^2}$$$$$$

Answered by A5T last updated on 30/Jun/24

$$y={tan}^{-1}\left(x\right)\Rightarrow tan\left(y\right)=x$$$$\frac{dx}{dy}={tan}^{2}y+1\Rightarrow \frac{dy}{dx}=\frac{1}{{tan}^{2}y+1}=\frac{1}{x^2+1}$$$$\Rightarrow \frac{d\left({tan}^{-1}\left(\frac{4x}{1+5x}\right)\right)}{dx}$$$$=\frac{1}{\frac{41x^2+1+10x}{{(1+5x)}^2}}\times (\frac{4(1+5x)}{{(1+5x)}^2}+\frac{-20x}{{(1+5x)}^2})$$$$=\frac{4}{41x^2+10x+1}$$$$Similarly,\frac{d\left({tan}^{-1}\left(\frac{2+3x}{3-2x}\right)\right)}{dx}$$$$=\frac{1}{\frac{13(1+x^2)}{{(3-2x)}^2}}\times \left(\frac{13}{{(3-2x)}^2}\right)=\frac{1}{1+x^2}$$$$\Rightarrow \frac{dy}{dx}=\frac{4}{41x^2+10x+1}+\frac{1}{1+x^2}$$

Commented by Spillover last updated on 01/Jul/24

$$$$How did you get that 41?

Commented by A5T last updated on 02/Jul/24

$$Ify={tan}^{-1}\left(x\right),then\frac{dy}{dx}=\frac{1}{x^2+1}$$$$y={tan}^{-1}\left[f\left(x\right)\right]\Rightarrow \frac{dy}{dx}=\frac{1}{{\left[f\left(x\right)\right]}^2+1}\times f^{\prime }\left(x\right)$$

Answered by Spillover last updated on 01/Jul/24

Commented by A5T last updated on 01/Jul/24

$$Thisisnotcorrect.$$

Commented by Spillover last updated on 01/Jul/24

$$why?$$

Commented by Spillover last updated on 01/Jul/24

$$recall$$$$\tan {}^{-1}\frac{4x}{1+5x}={\tan }^{-1}\frac{5x-x}{1+5x}$$$$$$

Commented by A5T last updated on 01/Jul/24

$$Haveyoutriedusingmultiplesoftwaresto$$$$calulatethis?$$$$Thisseemsdubious:$$$${tan}^{-1}\frac{5x-x}{1+5x}\stackrel{?}{=}{tan}^{-1}5x-{tan}^{-1}x$$

Commented by A5T last updated on 01/Jul/24

Commented by A5T last updated on 01/Jul/24

Commented by Spillover last updated on 01/Jul/24

Commented by A5T last updated on 01/Jul/24

$$Thiswouldgive:$$$${tan}^{-1}5x-{tan}^{-1}x={tan}^{-1}\left(\frac{5x-x}{1+\left(5x\right)x}\right)$$$$={tan}^{-1}\left(\frac{4x}{1+5x^2}\right)$$

Commented by Spillover last updated on 02/Jul/24

$$yourrightthankyouconfirmation$$$$ihadforgottenthatsquare$$

Commented by Spillover last updated on 02/Jul/24

$$nowiunderstandyouwhy$$$$mywayisdubious$$

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