6. (i) find $f(x)$ for $f(x)=(x^{2}+1)^{5}sin^{2}x$. (ii) find $g(x)$ for $g(x)=sqrt{\frac{sin(2pi x)}{1 +…

Explanation:

Step1: Apply product - rule for (f(x))

The product - rule states that if (y = u\cdot v), then (y'=u'v + uv'). Let (u=(x^{2}+1)^{5}) and (v = \sin^{2}x). First, find (u') using the chain - rule. If (y=(x^{2}+1)^{5}), let (t=x^{2}+1), then (y = t^{5}). (\frac{dy}{dt}=5t^{4}) and (\frac{dt}{dx}=2x), so (u' = 5(x^{2}+1)^{4}\cdot2x=10x(x^{2}+1)^{4}). To find (v'), since (v=\sin^{2}x), let (s = \sin x), then (v = s^{2}). (\frac{dv}{ds}=2s) and (\frac{ds}{dx}=\cos x), so (v'=2\sin x\cos x=\sin2x). Then (f'(x)=u'v + uv'=10x(x^{2}+1)^{4}\sin^{2}x+(x^{2}+1)^{5}\sin2x).

Step2: Apply quotient - rule and chain - rule for (g(x))

First, rewrite (g(x)=\left(\frac{\sin(2\pi x)}{1 + x^{2}}\right)^{\frac{1}{2}}). Let (y = g(x)), (u=\sin(2\pi x)), (v = 1 + x^{2}), and (w=\frac{u}{v}), so (y = w^{\frac{1}{2}}). By the chain - rule, (\frac{dy}{dw}=\frac{1}{2}w^{-\frac{1}{2}}). By the quotient - rule, (w'=\frac{u'v - uv'}{v^{2}}), where (u' = 2\pi\cos(2\pi x)) and (v' = 2x). So (w'=\frac{2\pi\cos(2\pi x)(1 + x^{2})-2x\sin(2\pi x)}{(1 + x^{2})^{2}}). Then (g'(x)=\frac{1}{2}\left(\frac{\sin(2\pi x)}{1 + x^{2}}\right)^{-\frac{1}{2}}\cdot\frac{2\pi\cos(2\pi x)(1 + x^{2})-2x\sin(2\pi x)}{(1 + x^{2})^{2}}).

Answer:

(i) (f'(x)=10x(x^{2}+1)^{4}\sin^{2}x+(x^{2}+1)^{5}\sin2x) (ii) (g'(x)=\frac{2\pi\cos(2\pi x)(1 + x^{2})-2x\sin(2\pi x)}{2(1 + x^{2})^{2}\sqrt{\frac{\sin(2\pi x)}{1 + x^{2}}}})