Given y - 5x- 1 - x 2 - cos2 2x -1 IIT-JEE- 1980find dydx-

Given y = 5x√( (1 x #160; )^2) + cos^2 (2x +1 ) #160; #160;Differentiating both sides w.r.t. x , we get dydx = 5 [ (1 x )^2/3 2/3 ...

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Cos(A+B)Cos(A-B)

Given y = 5...

Question

Given $y = \frac{5 x}{\sqrt[3]{{(1 - x)}^{2}}} + {cos}^{2} (2 x + 1)$ (IIT-JEE, 1980)
find $\frac{d y}{d x}$ .

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y = \frac{5 x}{{(1 - x)}^{2 / 3}} + {cos}^{2} (2 x + 1),

\frac{d y}{d x} .

y = \tan^{- 1} \{\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}\}, find \frac{d y}{d x}

\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)

\frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{1 - x^{2}}

y = \log \frac{x^{2} + x + 1}{x^{2} - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} x}{1 - x^{2}}), find \frac{d y}{d x} .

\frac{d y}{d x} = \frac{1 + y^{2}}{y^{3}}

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