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find dy/dx when y= log((x+sqrt(x^2+a^2))/(x-sqrt(x^2 – a^2)))

 
find dy/dx when
y=log((x+sqrt(x^2+a^2))/(x-sqrt(x^2 – a^2)))

Grade:12

1 Answers

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

y = log(\frac{x+\sqrt{x^{2}+a^{2}}}{x-\sqrt{x^{2}-a^{2}}})
Apply the chain rule
\frac{dy}{dx} = \frac{d(logu)}{dx}.\frac{du}{dx}
u = \frac{x+\sqrt{x^{2}+a^{2}}}{x-\sqrt{x^{2}-a^{2}}}
\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}
\frac{du}{dx} = \frac{(x-\sqrt{x^{2}+a^{2}}).(1+\frac{x}{\sqrt{x^{2}+a^{2}}})-(x+\sqrt{x^{2}+a^{2}}).(1-\frac{x}{\sqrt{x^{2}-a^{2}}})}{(x-\sqrt{x^{2}-a^{2}})^{2}}\frac{du}{dx} = \frac{(1+\frac{x}{\sqrt{x^{2}+a^{2}}})+(\frac{x+\sqrt{x^{2}+a^{2}}}{\sqrt{x^{2}-a^{2}}})}{(x-\sqrt{x^{2}-a^{2}})}
\frac{du}{dx} = \frac{(\frac{1}{\sqrt{x^{2}+a^{2}}})+(\frac{1}{\sqrt{x^{2}-a^{2}}})}{(x-\sqrt{x^{2}-a^{2}})}.(x+\sqrt{x^{2}+a^{2}})
\frac{du}{dx} = (\frac{1}{\sqrt{x^{2}+a^{2}}})+(\frac{1}{\sqrt{x^{2}-a^{2}}}).u
\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}
\frac{dy}{dx} = (\frac{1}{\sqrt{x^{2}+a^{2}}})+(\frac{1}{\sqrt{x^{2}-a^{2}}})




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