Math Doubts

Derivative of cscx formula Proof

The derivative of cosecant function with respect to a variable is equal to the negative product of cosecant and cotangent. If $x$ denotes a variable, then the cosecant function is expressed as $\csc{x}$ or $\operatorname{cosec}{x}$. The differentiation of the $\csc{x}$ function with respect to $x$ is equal to negative product of $\csc{x}$ and $\cot{x}$. The derivative of cosecant function is derived mathematically from first principle.

Differentiation of function in Limit form

As per definition of the derivative, write the derivative of a function in terms of $x$ in limits form.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

Now, take the change in $x$ is simply denoted by $h$. In other words, $\Delta x = h$.

$\implies$ $\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

If $f{(x)} = \csc{x}$, then $f{(x+h)} = \csc{(x+h)}$. Now, substitute them in the definition of the derivative to find the differentiation of $\csc{x}$ function with respect to $x$ by first principle.

$\implies$ $\dfrac{d}{dx}{\, (\csc{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\csc{(x+h)}-\csc{x}}{h}}$

Simplify the entire function

For simplifying the difference of the cosecant functions in the numerator, express each cosecant function in terms of sine function as per reciprocal identity of sin function.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1}{\sin{(x+h)}}-\dfrac{1}{\sin{x}}}{h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\sin{x}-\sin{(x+h)}}{\sin{(x+h)}\sin{x}}}{h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}-\sin{(x+h)}}{h\sin{(x+h)}\sin{x}}}$

In numerator, two sine functions are subtracting. The difference of them can be simplified by using difference of sine functions to product transforming trigonometric identity.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{x+x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-(x+h)}{2}\Bigg]}}{h\sin{(x+h)}\sin{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-x-h}{2}\Bigg]}}{h\sin{(x+h)}\sin{x}}}$

$=\,\,\,$ $\require{cancel} \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{\cancel{x}-\cancel{x}-h}{2}\Bigg]}}{h\sin{(x+h)}\sin{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{-h}{2}\Bigg]}}{h\sin{(x+h)}\sin{x}}}$

According to even/odd trigonometric identity of sin function, the sine of negative angle is equal to negative sin of angle.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\Bigg(-\sin{\Bigg[\dfrac{h}{2}\Bigg]\Bigg)}}{h\sin{(x+h)}\sin{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h\sin{(x+h)}\sin{x}}}$

Now, the trigonometric function has to be separated as product of two trigonometric functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{-\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\sin{(x+h)}\sin{x}} \times \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}\Bigg)}$

Use product rule of limits to find the limit of product of functions by the product of their limits.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\sin{(x+h)}\sin{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}}$

In second factor, the number $2$ multiplies the sine function in numerator and it divides the denominator. So, shift it to denominator.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\sin{(x+h)}\sin{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

Evaluate Limits of trigonometric functions

Firstly, evaluate the first factor by using direct substitution method and do not touch the second factor at this time.

$=\,\,\,$ $\dfrac{-\cos{\Bigg[\dfrac{2x+0}{2}\Bigg]}}{\sin{(x+0)}\sin{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{-\cos{\Bigg[\dfrac{2x}{2}\Bigg]}}{\sin{x}\sin{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\require{cancel} \dfrac{-\cos{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}}{\sin{x}\sin{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{-\cos{x}}{\sin{x}\sin{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\Bigg(-\dfrac{\cos{x}}{\sin{x}}\Bigg) \times \dfrac{1}{\sin{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

The quotient of cosine by sine function is equal to cotangent as per quotient identity of cos and sin functions. Similarly, the reciprocal of sine is cosecant as per reciprocal identity of sine function.

$=\,\,\,$ $(-\cot{x}) \times \csc{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $(-\cot{x}\csc{x})$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. So, $\dfrac{h}{2} \to 0$. Therefore, if $h$ approaches zero, then $\dfrac{h}{2}$ also tends to $0$.

$=\,\,\,$ $-\csc{x}\cot{x}$ $\times$ $\large \displaystyle \lim_{\frac{h}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

Take $y = \dfrac{h}{2}$ and write the limit of trigonometric function in terms of $y$.

$=\,\,\,$ $-\csc{x}\cot{x}$ $\times$ $\large \displaystyle \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}}$

According to limit of sinx/x as x approaches 0 formula, the limit of the trigonometric function is equal to $1$.

$=\,\,\,$ $-\csc{x}\cot{x} \times 1$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \csc{x}} \,=\, -\csc{x}\cot{x}$

Therefore, it is proved that the derivative of cosecant function is equal to the negative product of cosecant and cotangent functions.



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