Math Doubts

Derivative of cscx (or) cosecx 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 $\csc{x}$. 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 by first principle.

Express Differentiation of function in Limit form

The differentiation of function in terms of $x$ is written in the form of following limiting operation by the definition of the derivative.

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

If $f{(x)} = \csc{x}$, then $f{(x+h)} = \csc{(x+h)}$. The proof of derivative of $\csc{x}$ function with respect to $x$ can be started from 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

In order to simplify the difference of the cosecant functions in the numerator, each cosecant function should be expressed in sine function as per reciprocal identity of sin function.

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

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

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

Now, the difference of the sine functions can be combined by the difference to product identity of sin functions.

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

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

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

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

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 should be separated as product of trigonometric functions as follows.

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

The limit of product of functions is equal to product of their limits according to product rule of limits.

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

The number $2$ is a factor in the numerator of the second limiting function and it divides the denominator mathematically. So, it can be shifted to denominator for achieving second limiting function same as the limit of the trigonometric function.

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

Evaluate Limits of trigonometric functions

Now, the limit of each function can be evaluated but let’s do it one by one. So, evaluate the limit of first trigonometric function firstly by the direct substitution method.

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

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

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

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

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

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

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

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

Now, find the limit of second trigonometric function. It is almost similar to the limit of trigonometric function but the limiting input is not exactly same as the limit of $\sin{x}/{x}$ as $x$ approaches zero formula. Therefore, it essential to make this function same as the standard result.

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. So, $\dfrac{h}{2} \to 0$. It has proved that 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 $z = \dfrac{h}{2}$ and convert the limit of trigonometric function in terms of $z$.

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

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

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

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

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