Question

Differentiate the function w.r.t x:

Cos^-1 x/2 × (2x + 7)^-1/2

Answer 1

Given: f(x) =cos^-1\(\cfrac{\text x}2\) + (2x + 7)^-1/2

\(\cfrac{d}{d\text x}\)f(x) = cos^-1\(\cfrac{\text x}2\) x \(\cfrac{d}{d\text x}\)(2x + 7)^-1/2+ \(\left(\cfrac{d}{d\text x}cos^{-1}\cfrac{\text x}2\right)\)(2x + 7)^-1/2

\(\left(\because \cfrac{d}{d\text x}u\text v=u\times\cfrac{d\text v}{du}+\cfrac{du}{d\text x}\text v\right)\)

= \(-\cfrac{2}2\)(2x + 7)^-3/2cos^-1\(\cfrac{\text x}2\) - \(\cfrac{1}{\sqrt{1-\left(\frac{\text x}2\right)^2}}\times\)\(\cfrac12\) (2x + 7)^-1/2

(By chain rule)

 = -(2x + 7)^-3/2 cos^-1\(\cfrac{\text x}2\) - \(\cfrac{1}{\sqrt{4-\text x^2}}\) (2x + 7)^-1/2

= -(2x + 7)^-1/2\(\left(\cfrac{cos^{-1}\cfrac{\text x}2}{2\text x+7}+\cfrac1{\sqrt{4-\text x^2}}\right)\) 

= \(-\cfrac{1}{\sqrt{2\text x+ 7}}\) \(\left(\cfrac{cos^{-1}\cfrac{\text x}2}{2\text x+7}+\cfrac1{\sqrt{4-\text x^2}}\right)\)