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Question Number 24913 by chernoaguero@gmail.com last updated on 28/Nov/17

find the derivative of (sec(√(1+cos x)))

$$\mathrm{find}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of} \\ $$$$\left(\mathrm{sec}\sqrt{\left.\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)}\right. \\ $$

Commented by prakash jain last updated on 28/Nov/17

u=1+cos x v=(√(1+cos x)) (d/dx)sec ((√(1+cosx))) =(d/dx)sec v =(d/dv)sec v∙(d/dx)v =(d/dv)sec v∙(d/dx)(√u) =(d/dv)sec v∙(d/du)(√u)∙(d/dx)u =(d/dv)sec v∙(d/du)(√u)∙(d/dx)(1+cos x) =tan v∙sec v∙(1/(2(√u)))(−sin x) =tan (√(1+cos x))∙sec (√(1+cos x))∙(1/(2(√(1+cos x))))(−sin x)

$${u}=\mathrm{1}+\mathrm{cos}\:{x} \\ $$$${v}=\sqrt{\mathrm{1}+\mathrm{cos}\:{x}} \\ $$$$\frac{{d}}{{dx}}\mathrm{sec}\:\left(\sqrt{\mathrm{1}+\mathrm{cos}{x}}\right) \\ $$$$=\frac{{d}}{{dx}}\mathrm{sec}\:{v} \\ $$$$=\frac{{d}}{{dv}}\mathrm{sec}\:{v}\centerdot\frac{{d}}{{dx}}{v} \\ $$$$=\frac{{d}}{{dv}}\mathrm{sec}\:{v}\centerdot\frac{{d}}{{dx}}\sqrt{{u}} \\ $$$$=\frac{{d}}{{dv}}\mathrm{sec}\:{v}\centerdot\frac{{d}}{{du}}\sqrt{{u}}\centerdot\frac{{d}}{{dx}}{u} \\ $$$$=\frac{{d}}{{dv}}\mathrm{sec}\:{v}\centerdot\frac{{d}}{{du}}\sqrt{{u}}\centerdot\frac{{d}}{{dx}}\left(\mathrm{1}+\mathrm{cos}\:{x}\right) \\ $$$$=\mathrm{tan}\:{v}\centerdot\mathrm{sec}\:{v}\centerdot\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}\left(−\mathrm{sin}\:{x}\right) \\ $$$$=\mathrm{tan}\:\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\centerdot\mathrm{sec}\:\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\centerdot\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}}\left(−\mathrm{sin}\:{x}\right) \\ $$

Commented by chernoaguero@gmail.com last updated on 28/Nov/17

Thank alot sir

$$\mathrm{Thank}\:\mathrm{alot}\:\:\mathrm{sir} \\ $$$$ \\ $$

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