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Differential Calculus> diifferentiate sin^-1(2^(x+1) . 3^(x) / 1...

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diifferentiate sin^-1(2^(x+1) . 3^(x) / 1+(36)^x) with respect to x.

taniska , 11 Years ago

Grade 12

1 Answers

Arun Kumar

Hi

$\\\frac{d}{dx}\left(\arcsin \left(\frac{2^{x+1}\cdot \:3^x}{1+36^x}\right)\right) \\\mathrm{Let\:}\frac{2^{x+1}\cdot \:3^x}{1+36^x}\mathrm{=}u \\=\frac{d}{du}\left(\arcsin \left(u\right)\right)\frac{d}{dx}\left(\frac{2^{x+1}\cdot \:3^x}{1+36^x}\right) \\=\frac{1}{\sqrt{1-u^2}}\left(-\frac{2^{x+1}\cdot \:3^x\left(36^x-1\right)\ln \left(6\right)}{\left(36^x+1\right)^2}\right) \\\mathrm{Substitute}\:u=\frac{2^{x+1}\cdot \:3^x}{1+36^x} \\=\frac{1}{\sqrt{1-\left(\frac{2^{x+1}\cdot \:3^x}{1+36^x}\right)^2}}\left(-\frac{2^{x+1}\cdot \:3^x\left(36^x-1\right)\ln \left(6\right)}{\left(36^x+1\right)^2}\right) \\=-\frac{2^{x+1}\cdot \:3^x\ln \left(6\right)\text{sgn}\left(36^x-1\right)}{36^x+1} \\sgn(x)=\frac{x}{|x|}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Last Activity: 11 Years ago

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