IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Xavier4

cos^-1X/2 / root under (2X+7) where -2<X<2

versatile_boy

$y = \frac{{cos}^{-1}\frac{x}{2}}{\sqrt{2x+7}}$ .

Use Quotient rule

$\frac{dy}{dx} =\frac{\sqrt{2x+7}.\frac{-1}{\sqrt{1-\frac{{x}^{2}}{4}}}-\frac{{cos}^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7}$ $=\frac{-2\sqrt{\frac{2x+7}{4-{x}^{2}}}-\frac{{cos}^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7}$

Himanshu346

I have rated u

allamraju

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@versatile boy,again a small mistake,You didn't write (1/2) in the derivative of cos^-1(x/2) which gets cancelled with 2.Hence,the answer is

.

Ask & Discuss Questions with Community & Experts

@versatile boy,again a small mistake,You didn't write (1/2) in the derivative of cos-1(x/2) which gets cancelled with 2.Hence,the answer is

.

@versatile boy,again a small mistake,You didn't write (1/2) in the derivative of cos^-1(x/2) which gets cancelled with 2.Hence,the answer is