What is the equation of the line tangent to f(x)=cscx at x=pi/2 ?

1 Answer
Jan 14, 2016

y=1

Explanation:

Given:

y=f(x)

The equation of the tangent line in x=x_0 is:

(y-f(x_0))=f'(x_0)(x-x_0)

with f'(x_o)=m slope of the line

f(x)=csc(x)=1/sin(x)

f'(x)=d/dx(csc(x))=d/dx(1/sin(x))

Using the Quotient Rule

d/dx(h(x))=d/dx(g(x)/(i(x)))=(g'(x)*i(x)-i'(x)*g(x))/(i^2(x))

:. f'(x)=(0*sinx-(cosx)*1)/sin^2(x)=-cos(x)/sin^2(x)

f'(pi/2)=-0/1=0

f(pi/2)=csc(pi/2)=1

insert the values in the tangent line equation:

:. y-1=0(x-pi/2)

y-1=0

y=1

Alternatively, without the derivation, if you remember from trigonometric that:

y=csc(x) has a local minimum point in

x_0=pi/2

then you can affirm that the tangent line is

y=f(x_0)