Which of the following best explains why the slope of the graph of y=tanx increases without bound as x approaches 2π from the left?
a) Because cosx approaches 0+, the reciprocal secx grows without bound; since the derivative is secx, the slope diverges.
b) Because sinx approaches 1, the derivative sin2x approaches 1, so the slope remains finite.
c) Because cosx approaches 0+, the expression sec2x=cos2x1 grows without bound, so the derivative diverges.
d) Since cosx changes sign near 2π, the factor sec2x becomes negative, causing the derivative to approach −∞.