JEE Main 2024 — Differentiation Question with Solution

Given:$y=\frac{\left(\sqrt{x}+1\right)\left(x^2-\sqrt{x}\right)}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=\frac{\left(\sqrt{x}+1\right)\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(\sqrt{x}-1\right)}+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=\frac{\left(\sqrt{x}+1\right)\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=\frac{\left(\sqrt{x}+1\right)\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left({\left(\sqrt{x}\right)}^3-1\right)}+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left({\left(\sqrt{x}\right)}^3-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left({\left(\sqrt{x}\right)}^3-1\right)}+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow y=x-1+\frac{1}{15}\left(3{\cos }^{5}x-5{\cos }^{3}x\right)$

$\Rightarrow \frac{dy}{dx}=1+\frac{1}{15}\left(15{\cos }^{4}x\left(-\sin x\right)+15{\cos }^{2}x\sin x\right)$

$\Rightarrow \frac{dy}{dx}=1+\left({\cos }^{4}x\left(-\sin x\right)+{\cos }^{2}x\sin x\right)$

$\Rightarrow y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=1+\left({\cos }^4\left(\frac{\mathrm{\pi }}{6}\right)\left(-\sin \left(\frac{\mathrm{\pi }}{6}\right)\right)+{\cos }^2\left(\frac{\mathrm{\pi }}{6}\right)\sin \left(\frac{\mathrm{\pi }}{6}\right)\right)$

$\Rightarrow y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=1+\left(\frac{{\left(\sqrt{3}\right)}^4}{2^4}\left(-\frac{1}{2}\right)+\frac{{\left(\sqrt{3}\right)}^2}{2^2}\times \frac{1}{2}\right)$

$\Rightarrow y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=1-\frac{9}{32}+\frac{3}{8}$

$\Rightarrow y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{32-9+12}{32}$

$\Rightarrow y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{35}{32}$

$\Rightarrow 96y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=35\times 3$

$\Rightarrow 96y^{\text{'}}\left(\frac{\mathrm{\pi }}{6}\right)=105$