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find dy/dx when y= s in^-1(2^(x+1)/ 1+ (4)^x)

 
find dy/dx when
y= sin^-1(2^(x+1)/ 1+ (4)^x)

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find answer to your question below

y = sin^{-1}(\frac{2^{x+1}}{4^{x}+1})
Using the chain rule
\frac{dy}{dx} = \frac{d(sin^{-1}(\frac{2^{x+1}}{4^{x}+1}))}{dx} = \frac{d(sin^{-1}u)}{dx}.\frac{du}{dx}
u = \frac{2^{x+1}}{4^{x}+1}
\frac{dy}{dx} = \frac{1}{\sqrt{1-u^{2}}}.\frac{du}{dx}
\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{2^{x+1}}{4^{x}+1})^{2}}}.\frac{(4^{x}+1).2^{x+1}.log2-2^{x+1}.(4^{x})log4}{(4^{x}+1)^2}
\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{2^{x+1}}{4^{x}+1})^{2}}}.\frac{(4^{x}+1).2^{x+1}.log2-2^{x+1}.(4^{x})2log2}{(4^{x}+1)^2}
\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{2^{x+1}}{4^{x}+1})^{2}}}.\frac{log2(2^{x+1})[(4^{x}+1)-2(4^{x})]}{(4^{x}+1)^2}
\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{2^{x+1}}{4^{x}+1})^{2}}}.\frac{log2(2^{x+1})(1-4^{x})}{(4^{x}+1)^2}
\frac{dy}{dx} = \frac{1}{\sqrt{(\frac{4^{x}-1}{4^{x}+1})^{2}}}.\frac{log2(2^{x+1})(1-4^{x})}{(4^{x}+1)^2}
\frac{dy}{dx} = \frac{1}{|4^{x}-1|}.\frac{log2(2^{x+1})(1-4^{x})}{(4^{x}+1)}

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