1

I want to find out the derivative of 1/sin(x) without using the reciprocal rule. Let f(x) = 1/sin(x) Df/dx = (f(x+h) - f(x))/ h

I keep getting 0 as the answer while the actual derivative according to the reciprocal rule is -cos(x)/sin^2(x)

Can anybody help me with this?

Minu
  • 971
  • 1
  • 9
  • 16

2 Answers2

2

Assume $\sin(x)\neq0$ then we compute the derivative as follows:

$\displaystyle{\frac{\frac{1}{\sin(x+h)}-\frac{1}{\sin(x)}}{h}=\frac{\frac{\sin(x)-\sin(x+h)}{\sin(x)\sin(x+h)}}{h}=\frac{\sin(x)-(\sin(x)\cos(h)+\sin(h)\cos(x))}{h\sin(x)\sin(x+h)}=\frac{\sin(x)(1-\cos(h))-\sin(h)\cos(x)}{h\sin(x)\sin(x+h)}}$

$=\displaystyle{\frac{1-\cos(h)}{h}\cdot\frac{1}{\sin(x+h)}-\frac{\sin(h)}{h}\frac{\cos(x)}{\sin(x)\sin(x+h)}=\frac{1-\cos(h)}{h^{2}}\cdot h\cdot\frac{1}{\sin(x+h)}-\frac{\sin(h)}{h}\cdot\frac{\cos(x)}{\sin(x)\sin(x+h)}}$

You should now be able to evaluate the limit.

Cure
  • 4,051
user71352
  • 13,038
  • What happens if I don't take sinx common in the numerator and if I expand sin(x+h) in the denominator. Logically, the result should be the same after evaluating the limits. – Minu Jun 25 '14 at 23:43
1

Let $y = \frac1{\sin x}$, so that $$ y\sin x = 1 $$ Differentiating implicitly wrt $x$, we get $$ y'\sin x + y\cos x = 0 $$ Now replace $y$ with $\frac1{\sin x}$ and solve for $y'$.