Sunday, November 20, 2016

Calculus: Early Transcendentals, Chapter 3, 3.3, Section 3.3, Problem 14

y = (1 - sec(x))/tan(x)
=(1 /tan(x))- (sec(x)/tan(x))
y= cot(x) - csc(x)
so now the y' is as follows
y' = d/(dx) (cot(x) - csc(x))
= d/(dx) (cot(x)) - d/(dx)(csc(x))

as
d/(dx) (cot(x))
=d/(dx) (cos(x)/sin(x))
= ( ((cosx *(d/(dx) sin x)) -((d/(dx) cos x * sinx)))/(sin^2 x)
= (-cos^2 x -sin^2 x)/(sin^2 x)
= - 1/(sin^2 x)
= -csc^2(x)
and
d/(dx) (csc(x))
=d/(dx) (1/sin(x))
= ((d/(dx) sin x)*1 - sinx * d/dx(1))/(sin^2(x))
= -cosx/sin^2(x)
=-csc x cot(x)
so,
y' =d/(dx) (cot(x)) -d/(dx)(csc(x))
=-csc^2(x)-csc x cot(x)

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