JEE Advance - Mathematics (1980 - No. 23)
Given $$y = {{5x} \over {3\sqrt {{{\left( {1 - x} \right)}^2}} }} + {\cos ^2}\left( {2x + 1} \right)$$; Find $${{dy} \over {dx}}$$.
$$\frac{dy}{dx} = \frac{5}{3} \cdot \frac{1}{(1-x)^2} - 2\sin(4x+2), x < 1$$ and $$\frac{dy}{dx} = -\frac{5}{3} \cdot \frac{1}{(x-1)^2} - 2\sin(4x+2), x > 1$$
$$\frac{dy}{dx} = \frac{5}{3} \cdot \frac{1}{(1-x)^2} + 2\sin(4x+2)$$
$$\frac{dy}{dx} = \frac{5}{3} \cdot \frac{1}{1-x} - 2\sin(4x+2)$$
$$\frac{dy}{dx} = -\frac{5}{3} \cdot \frac{1}{(x-1)^2} - 2\sin(4x+2)$$
$$\frac{dy}{dx} = \frac{5}{3} \cdot \frac{1}{1-x} + 2\sin(4x+2)$$
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