1. find the tangent to ( y = x sin ^ { 2 } ( pi x ) ) at ( x = 1 / 2 ).

Explanation:

Step1: Find the value of (y) at (x = \frac{1}{2})

Substitute (x=\frac{1}{2}) into (y = x\sin^{2}(\pi x)). [ \begin{align*} y&=\frac{1}{2}\sin^{2}\left(\pi\times\frac{1}{2}\right)\ &=\frac{1}{2}\times1^{2}\ &=\frac{1}{2} \end{align*} ]

Step2: Differentiate (y) using the product rule ((uv)^\prime = u^\prime v+uv^\prime)

Let (u = x) and (v=\sin^{2}(\pi x)). First, find (u^\prime) and (v^\prime). (u^\prime=1). For (v=\sin^{2}(\pi x)), use the chain - rule. Let (t = \sin(\pi x)), then (v=t^{2}). (\frac{dv}{dt} = 2t) and (\frac{dt}{dx}=\pi\cos(\pi x)). So (v^\prime = 2\sin(\pi x)\times\pi\cos(\pi x)=\pi\sin(2\pi x)) (using (\sin(2\alpha)=2\sin\alpha\cos\alpha)). Then (y^\prime=\sin^{2}(\pi x)+x\times\pi\sin(2\pi x))

Step3: Find the slope (m) of the tangent at (x = \frac{1}{2})

Substitute (x=\frac{1}{2}) into (y^\prime). [ \begin{align*} y^\prime\left(\frac{1}{2}\right)&=\sin^{2}\left(\pi\times\frac{1}{2}\right)+\frac{1}{2}\times\pi\sin\left(2\pi\times\frac{1}{2}\right)\ &=1^{2}+\frac{\pi}{2}\times0\ &= 1 \end{align*} ]

Step4: Use the point - slope form (y - y_{0}=m(x - x_{0}))

Here (x_{0}=\frac{1}{2}), (y_{0}=\frac{1}{2}) and (m = 1). [ \begin{align*} y-\frac{1}{2}&=1\times\left(x-\frac{1}{2}\right)\ y-\frac{1}{2}&=x-\frac{1}{2}\ y&=x \end{align*} ]

Answer:

The equation of the tangent line is (y = x)