Answer:
[tex](\frac{dy}{dx})x_{= \frac{\pi }{2} } = f^{1} (\frac{\pi }{2} ) =2[/tex]
Step-by-step explanation:
Step(i):-
Given that
[tex]f(x) = \frac{5x}{sinx} -3xsinx[/tex]
Apply (UV )¹ = UV¹+VU¹
[tex]\frac{d}{dx} (\frac{U}{V} ) = \frac{V U^{l} -VU^{l} }{V^{2} }[/tex]
Step(ii):-
[tex]f^{1} (x) = 5\frac{sin x(1)-x(cos x)}{sin^{2}x } - 3(x cos x + sin x(1))[/tex]
put [tex]x = \frac{\pi }{2}[/tex]
[tex]f^{1} (\frac{\pi }{2} ) = 5\frac{sin \frac{\pi }{2} (1)-\frac{\pi }{2} (cos \frac{\pi }{2} )}{sin^{2}(\frac{\pi }{2} ) } - 3(\frac{\pi }{2} cos \frac{\pi }{2} + sin(\frac{\pi }{2}) (1))[/tex]
we know that
[tex]cos(\frac{\pi }{2}) = 0[/tex]
[tex]sin(\frac{\pi }{2} ) = 1[/tex]
[tex]f^{l} (\frac{\pi }{2} ) = \frac{5(1-0)}{1} -3(0+1)[/tex]
[tex]f^{1} (\frac{\pi }{2} ) = 5-3 =2[/tex]
Final answer:-
[tex](\frac{dy}{dx})x_{= \frac{\pi }{2} } = f^{1} (\frac{\pi }{2} ) =2[/tex]
