Question

F(x)= (5x/sinx) - 3xsinx Solve for dy/dx at x = Pi over 2

Answer 1

Answer:

$(\frac{dy}{dx})x_{= \frac{\pi }{2} } = f^{1} (\frac{\pi }{2} ) =2$

Step-by-step explanation:

Step(i):-

Given that

 $f(x) = \frac{5x}{sinx} -3xsinx$

Apply (UV )¹ = UV¹+VU¹

$\frac{d}{dx} (\frac{U}{V} ) = \frac{V U^{l} -VU^{l} }{V^{2} }$

Step(ii):-

$f^{1} (x) = 5\frac{sin x(1)-x(cos x)}{sin^{2}x } - 3(x cos x + sin x(1))$

put  $x = \frac{\pi }{2}$

$f^{1} (\frac{\pi }{2} ) = 5\frac{sin \frac{\pi }{2} (1)-\frac{\pi }{2} (cos \frac{\pi }{2} )}{sin^{2}(\frac{\pi }{2} ) } - 3(\frac{\pi }{2} cos \frac{\pi }{2} + sin(\frac{\pi }{2}) (1))$

we know that  

$cos(\frac{\pi }{2}) = 0$  

$sin(\frac{\pi }{2} ) = 1$

$f^{l} (\frac{\pi }{2} ) = \frac{5(1-0)}{1} -3(0+1)$

$f^{1} (\frac{\pi }{2} ) = 5-3 =2$

Final answer:-

$(\frac{dy}{dx})x_{= \frac{\pi }{2} } = f^{1} (\frac{\pi }{2} ) =2$