If y=(x+1)(x^2-x)xx+x+x+(1)/(15)(3 ^2x-5) ^3x, then 96y^'(( )/(6)) is equal to : Answer: 1…

Question

If $y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x$, then $96{y}^{'}(\frac{\pi }{6})$ is equal to :

Answer: 105

Accepted Answer

$\begin{gathered} y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x \ y=\frac{(\sqrt{x}+1)(\sqrt{x})((\sqrt{x}{)}^{3}-1)}{(\sqrt{x})((\sqrt{x}{)}^{2}+(\sqrt{x})+1)}+\frac{1}{5}{\cos }^{5}x-\frac{1}{3}{\cos }^{3}x \ y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5}{\cos }^{5}x-\frac{1}{3}{\cos }^{3}x \end{gathered}$

$\begin{gathered} {y}^{'}=1-{\cos }^{4}x\cdot (\sin x)+{\cos }^{2}x(\sin x \ {y}^{'}(\frac{\pi }{6})=1-\frac{9}{16}	imes \frac{1}{2}+\frac{3}{4}	imes \frac{1}{2} \ =\frac{32-9+12}{32}=\frac{35}{32} \ 	herefore 96{y}^{'}(\frac{\pi }{6})=105 \end{gathered}$

Solutions · 1