Find the derivative of #cscx# from first principles?

Question

Isaiah Allen · Accepted Answer

#(dy)/(dx)=-cscxcotx# As #y=cscx=1/sinx#, #y+deltay=1/sin(x+deltax)# Hence #deltay=1/sin(x+deltax)-1/sinx# = #(sinx-sin(x+deltax))/(sinxsin(x+deltax))# = #-(sin(x+deltax)-sinx)/(sinxsin(x+deltax))# = #-(2sin((x+deltax-x)/2)cos((x+deltax+x)/2))/(sinxsin(x+deltax))# = #-(2sin((deltax)/2)cos(x+(deltax)/2))/(sinxsin(x+deltax))# and #(deltay)/(deltax)=-(2sin((deltax)/2)cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))# Hence #(dy)/(dx)=Lt_(deltax->0)(deltay)/(deltax)# = #Lt_(deltax->0)-(2sin((deltax)/2)cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))# = #Lt_(deltax->0)(-sin((deltax)/2)/((deltax)/2))xxLt_(deltax->0)[(cos(x+(deltax)/2))/(deltaxsinxsin(x+deltax))]# = #-1xxcosx/(sin^2x)=-1/(sinx)xxcosx/sinx=-cscxcotx#

William Abdalla · Answer

undefined To find the derivative of $ \csc(x) $ from first principles, we start with the definition of the cosecant function:

$$ \csc(x) = \frac{1}{\sin(x)} $$

Then, we use the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substituting $ f(x) = \csc(x) = \frac{1}{\sin(x)} $ into the definition of the derivative, we have:

$$ \csc'(x) = \lim_{h \to 0} \frac{\frac{1}{\sin(x + h)} - \frac{1}{\sin(x)}}{h} $$

To simplify, we can find a common denominator:

$$ \csc'(x) = \lim_{h \to 0} \frac{\sin(x) - \sin(x + h)}{h \cdot \sin(x) \cdot \sin(x + h)} $$

Using the trigonometric identity $ \sin(a) - \sin(b) = 2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) $, we simplify further:

$$ \csc'(x) = \lim_{h \to 0} \frac{2\cos\left(\frac{x+(x+h)}{2}\right)\sin\left(\frac{x-(x+h)}{2}\right)}{h \cdot \sin(x) \cdot \sin(x + h)} $$

$$ \csc'(x) = \lim_{h \to 0} \frac{2\cos\left(x + \frac{h}{2}\right)\sin\left(-\frac{h}{2}\right)}{h \cdot \sin(x) \cdot \sin(x + h)} $$

$$ \csc'(x) = \lim_{h \to 0} \frac{-2\cos\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h \cdot \sin(x) \cdot \sin(x + h)} $$

$$ \csc'(x) = \lim_{h \to 0} \frac{-2\cos\left(x + \frac{h}{2}\right)}{\sin(x) \cdot \sin\left(x + \frac{h}{2}\right)} $$

Now, as $ h \to 0 $, $ \frac{h}{2} \to 0 $, so:

$$ \lim_{h \to 0} \sin\left(x + \frac{h}{2}\right) = \sin(x) $$

$$ \csc'(x) = \lim_{h \to 0} \frac{-2\cos\left(x + \frac{h}{2}\right)}{\sin^2(x)} $$

$$ \csc'(x) = \frac{-2\cos(x)}{\sin^2(x)} $$

Thus, the derivative of $ \csc(x) $ from first principles is $ \csc'(x) = \frac{-2\cos(x)}{\sin^2(x)} $.