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?
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $y'$.
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