Question

What is the equation of the line tangent to `f(x)=x^2 + sin2x` at `x=pi/4`?

Answer 1

`y=(\pix)/2+(16-\pi^2)/16`

Explanation:

`y=x^2+sin2x=u+v`

`(dy)/(dx)=uv'+vu'`

`u'=2x`

`v'=2cos2x`

`(dy)/(dx)=2xsin2x+2x^2cos2x`

`2(\pi/4)sin(2*\pi/4)+2(\pi/4)^2cos(2*\pi/4)=\pi/2`

`y=(\pi/4)^2+sin(2*\pi/4)=1+\pi^2/16`

`1/1+\pi^2/16=(16+\pi^2)/16`

`y=mx+c`
`(16+\pi^2)/16=\pi/2(\pi/4)+c`

`c=(16+\pi^2)/16-\pi^2/8=(16+\pi^2-2\pi^2)/16=(16-\pi^2)/16`

`y=(\pix)/2+(16-\pi^2)/16`