Application Of Derivatives

Question

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Differentiate $\tan^{- 1} (\frac{1 + \cos x}{\sin x})$ with respect x.

Solution

${Let f(x) = tan}^{- 1} (1 + \frac{\cos x}{\sin x}) ∴ f' (x) = \frac{1}{1 + {(\frac{1 + \cos x}{\sin x})}^{2}} x \frac{d}{dx} (\frac{1 + \cos x}{\sin x}) = \frac{\sin^{2} x}{\sin^{2} x + (1 + \cos x)^{2}} x \frac{\sin x (- sinx) - (1 + \cos x) cosx}{\sin^{2} x} = \frac{\sin^{2} x}{\sin^{2} x + 1 + \cos^{2} x + 2 \cos x} x \frac{(- \sin^{2} x - \cos x - \cos^{2} x)}{\sin^{2} x} = - \frac{\cos x - 1}{1 + 1 + 2 \cos x} = \frac{- (1 + \cos x)}{2 (1 + \cos x)} = \frac{- 1}{2}$

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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Application Of Derivatives

Differentiate $\tan^{- 1} (\frac{1 + \cos x}{\sin x})$ with respect x.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Differentiate tan-1 1 + cos xsin x with respect x.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Differentiate $\tan^{- 1} (\frac{1 + \cos x}{\sin x})$ with respect x.