Skip to main content
Back to Differentiation
JEE Main 2024
Differentiation
Differentiation
Medium

Question

If y=(x+1)(x2x)xx+x+x+115(3cos2x5)cos3xy=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x, then 96y(π6)96 y^{\prime}\left(\frac{\pi}{6}\right) is equal to :

Answer: 1

Solution

y=(x+1)(x2x)xx+x+x+115(3cos2x5)cos3xy=(x+1)(x)((x)31)(x)((x)2+(x)+1)+15cos5x13cos3xy=(x+1)(x1)+15cos5x13cos3x\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x\end{aligned} y=1cos4x(sinx)+cos2x(sinxy(π6)=1916×12+34×12=329+1232=353296y(π6)=105\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}

Practice More Differentiation Questions

View All Questions