Derivative of x/sinx using first principle?

1 Answer
Mar 1, 2018

f'(x)=csc(x)-xcot(x)csc(x)

Explanation:

We want differentiate f(x)=x/sin(x), therefore we seek

f'(x)=lim_(h->0)((x+h)/sin(x+h)-x/sin(x))/h

For later use remember the two quite fundamental limits

color(orange)(lim_(x->0)sin(x)/x=1) and color(green)(lim_(x->0)(1-cos(x))/x=0)

and the trigonometric identity

color(brown)(sin(x+h)=cos(x)sin(h)+cos(h)sin(x))

Let's rewrite our problem

f'(x)=lim_(h->0)((x+h)/(sin(x+h))-x/sin(x))/h

=lim_(h->0)((x+h)sin(x)-xsin(x+h))/(hsin(x+h)sin(x))

=lim_(h->0)(hsin(x))/(hsin(x+h)sin(x))+lim_(h->0)(xsin(x)-xsin(x+h))/(hsin(x+h)sin(x))

=L_1+L_2

Let's rewrite the second limit

L_2=lim_(h->0)(xsin(x)-xcolor(brown)((cos(x)sin(h)+cos(h)sin(x))))/(hsin(x+h)sin(x))

=lim_(h->0)(xsin(x)(1-cos(h))-x(cos(x)sin(h)))/(hsin(x+h)sin(x))

=lim_(h->0)(xsin(x)(1-cos(h)))/(hsin(x+h)sin(x))-lim_(h->0)(xcos(x)sin(h))/(hsin(x+h)sin(x))

=L_3-L_4

Such that f'(x)=L_1+L_3-L_4

What is left to evaluate these limits

Limit 1

L_1=lim_(h->0)(hsin(x))/(hsin(x+h)sin(x))=csc(x)

Limit 3

L_3=lim_(h->0)(xsin(x)(1-cos(h)))/(hsin(x+h)sin(x))

=xcolor(green)(lim_(h->0)(1-cos(h))/(h))lim_(h->0)1/sin(x+h)=0

Limit 4

L_4=lim_(h->0)(xcos(x)sin(h))/(hsin(x+h)sin(x))

=xcos(x)/sin(x)color(orange)(lim_(h->0)(sin(h))/(h))lim_(h->0)1/sin(x+h)=xcot(x)csc(x)

Remember f'(x)=L_1+L_3-L_4 so

f'(x)=csc(x)-xcot(x)csc(x)