Mathematics · Methods of DifferentiationJEE Main 2024 — 1 February, Shift 2 — Question 24If y=(x+1)(x2−x)xx+x+x+115(3cos2x−5)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} xy=xx+x+x(x+1)(x2−x)+151(3cos2x−5)cos3x then 96y′(π6)96 y^{\prime}\left(\frac{\pi}{6}\right)96y′(6π) is equal to :Answer: 105Numerical answer — enter this value.Step-by-step solutiony=(x+1)(x2−x)xx+x+x+115(3cos2x−5)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} xy=xx+x+x(x+1)(x2−x)+151(3cos2x−5)cos3x y=(x+1)(x)((x)3−1)(x)((x)2+(x)+1)+15cos5x−13cos3xy=\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} xy=(x)((x)2+(x)+1)(x+1)(x)((x)3−1)+51cos5x−31cos3x y=(x+1)(x−1)+15cos5x−13cos3x\mathrm{y}=(\sqrt{\mathrm{x}}+1)(\sqrt{\mathrm{x}}-1)+\frac{1}{5} \cos ^{5} \mathrm{x}-\frac{1}{3} \cos ^{3} \mathrm{x}y=(x+1)(x−1)+51cos5x−31cos3x y′=1−cos4x⋅(sinx)+cos2x(sinx)y^{\prime}=1-\cos ^{4} x \cdot(\sin x)+\cos ^{2} x(\sin x)y′=1−cos4x⋅(sinx)+cos2x(sinx) y′(π6)=1−916×12+34×12y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2}y′(6π)=1−169×21+43×21 =32−9+1232=3532=\frac{32-9+12}{32}=\frac{35}{32}=3232−9+12=3235 =96y′(π6)=105=96 \mathrm{y}^{\prime}\left(\frac{\pi}{6}\right)=105=96y′(6π)=105Answer key and solution verified before publishing.Practise Methods of DifferentiationStart with this question, then two more from the same chapter — with a tutor that explains every step. Free.Solve a similar one free→ExamJEE Main 2024Paper1 February, Shift 2SubjectMathematicsChapterMethods of DifferentiationTopicMethods of Differentiation← Question 23Let f:(0, infty) arrow R and F(x)=int 0^x tf(t) d t . If F (x^2 )= x^4+x^5 , then sum r=1^12 f (r^2 ) is equal to:Question 25 →Let veca=hati+hatj+hatk, vecb=-hati-8 hatj+2 hatk and overrightarrow c=4 hat i+ c 2 hat j+ c 3 hat k be three vectors such that vecb ×…