<span>y = tan^−1(x2/4)</span>
tan(y) = x2/4
sec2(y) = x/2
y′ = xcos^2(y)/2
<span>cos^2(y) = <span>16x2+16</span></span>
<span>y′ = <span>8x/(<span>x2+16)
let u be x2+16
du is 2x dx
dy = 4 du / u
y = 4 ln (</span></span></span>x2 <span>+ 16)
y at x =0 = </span> 4 ln (<span>16) = 11.09</span>
Remark
I would have had the answer a whole lot sooner if I would have read the question properly. The figure in the circle is called a cyclic quadrilateral. It has the odd property that the angles that are opposite each other add up to 180o.
So DEB + DCB = 180o
DEB = 180 - 87
DEB = 93o
Note: The arcs marked 60 and 76 have nothing whatever to do with this problem.
Answer:
Mixed numbers have a whole number followed by the fraction (2 1/2). You would say "two and one half." The other format is an improper fraction where the numerator is greater than the denominator (5/2). Mathematicians would say that is five halves.
