Question

Find dy/dx

Answer 1

\[\LARGE ~~y=\sqrt{x^2+1}-\log(\frac{1}{x}+\sqrt{x+\frac{1}{x}})\]

Answer 2

\[\frac{dy}{dx}=\frac{1}{\cancel{2} \sqrt{x^2+1}} \times \cancel{2}x - \frac{1}{{ \frac{1}{x}}+\sqrt{x+\frac{1}{x}}} \times (\frac{-1}{x^2} + \frac{1}{2 \sqrt{ x + \frac{1}{x}}} \times (1 - \frac{1}{x^2}))\]
i got this..how do i simplify further?

Answer 3

http://www.wolframalpha.com/input/?i=y%3D+sqrt%28x%5E2%2B1%29+-+log+%281%2Fx+%2B+sqrt%28x+%2B1%2Fx%29

Answer 4

using the power rule alongwith logrithmic differentiation
if
\[\Large y=x^n\]
\[\Large \frac{dy}{dx}=nx^{n-1}\]
and
\[\Large y=\log (f(x)\]
\[\Large \frac{dy}{dx}=\frac{1}{\ln(10) f(x)}*f'(x)\]

using the above rules with chain rule
\[\Large \frac{dy}{dx}=\frac{1}{2}(x^2+1)^{\frac{1}{2}-1}*\frac{d}{dx(x^2+1)}-\frac{1}{\ln(10)(\frac{1}{x}+\sqrt{x+\frac{1}{x}}}*\frac{d}{dx}(\frac{1}{x}+\sqrt{x+\frac{1}{x}}\]