assignment 5: problem 11\n(1 point)\nlet ( f(x)=\frac{9 sin x}{4 sin x+6 cos x} ).\nthen ( f^{prime}(x)=…

Explanation:

Step1: Find the derivative using the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y'=\frac{u'v - uv'}{v^{2}}). Here, (u = 9\sin x), (u'=9\cos x), (v = 4\sin x+6\cos x), (v'=4\cos x - 6\sin x). [ \begin{align*} f'(x)&=\frac{9\cos x(4\sin x + 6\cos x)-9\sin x(4\cos x - 6\sin x)}{(4\sin x + 6\cos x)^{2}}\ &=\frac{36\sin x\cos x+54\cos^{2}x - 36\sin x\cos x + 54\sin^{2}x}{(4\sin x + 6\cos x)^{2}}\ &=\frac{54(\sin^{2}x+\cos^{2}x)}{(4\sin x + 6\cos x)^{2}} \end{align*} ] Since (\sin^{2}x+\cos^{2}x = 1), (f'(x)=\frac{54}{(4\sin x + 6\cos x)^{2}})

Step2: Find (m) (the slope of the tangent line at (x = 0))

Substitute (x = 0) into (f'(x)). When (x = 0), (\sin(0)=0) and (\cos(0)=1) [ m=f'(0)=\frac{54}{(4\times0 + 6\times1)^{2}}=\frac{54}{36}=\frac{3}{2} ]

Step3: Find (b) (the (y) - intercept)

First, find (f(0)). (f(0)=\frac{9\sin(0)}{4\sin(0)+6\cos(0)} = 0) The equation of the tangent line is (y=mx + b). Substituting (x = 0), (y = 0) and (m=\frac{3}{2}) into (y=mx + b), we get (0=\frac{3}{2}\times0 + b), so (b = 0)

Answer:

(f'(x)=\frac{54}{(4\sin x + 6\cos x)^{2}}), (m=\frac{3}{2}), (b = 0)