Differentiation Question 1
Question 1 - 2024 (01 Feb Shift 2)
If $y=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^{2} x-5\right) \cos ^{3} x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to :
Show Answer
Answer (105)
Solution
$y=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^{2} x-5\right) \cos ^{3} x$
$y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^{3}-1\right)}{(\sqrt{x})\left((\sqrt{x})^{2}+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^{5} x-\frac{1}{3} \cos ^{3} x$
$y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^{5} x-\frac{1}{3} \cos ^{3} x$
$y^{\prime}=1-\cos ^{4} x \cdot(\sin x)+\cos ^{2} x(\sin x)$
$y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2}$
$=\frac{32-9+12}{32}=\frac{35}{32}$
$=96 y^{\prime}\left(\frac{\pi}{6}\right)=105$
