Differentiate sin-1x w r t log( 1+x )

Let u = sin-1xand v = log1+xNow dudx=11-x2and dvdx=11+xSo dudv=dudxdvdx=11-x211+x=x+11-x2=x+121-x1+x=x+11-x

Differentiate sin-¹x w.r.t log( 1+x )

Let u = \sin^{- 1} x and v = \log (1 + x) Now \frac{du}{dx} = \frac{1}{\sqrt{1 - x^{2}}} and \frac{dv}{dx} = \frac{1}{1 + x} So \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{\sqrt{1 - x^{2}}}}{\frac{1}{1 + x}} = \frac{x + 1}{\sqrt{1 - x^{2}}} = \sqrt{\frac{{(x + 1)}^{2}}{(1 - x) (1 + x)}} = \sqrt{\frac{x + 1}{1 - x}}