Math Doubts

Derivative of cscx (or) cosecx formula Proof

Literal $x$ is taken as an angle of the right angled triangle and the cosecant function is written as $\csc{x}$ or $\operatorname{cosec}{x}$ in mathematics. The derivative of cosecant function with respect to $x$ is expressed as $\dfrac{d}{dx} \csc{x}$ or $\dfrac{d}{dx} \operatorname{cosec}{x}$ in mathematics.

Express Differentiation of function in Limit form

The differentiation of a function can be expressed in limit form as per the relation of derivative of a function in limit form.

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

Take $f(x) = \csc{x}$, then $f(x) = \csc{(x+h)}$. Now, substitute them in the differentiation rule of the function in limit form.

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

Transform cosecant functions in terms of sin functions

In trigonometry, there are no trigonometric identities with cosecant function. So, each cosecant function should be converted in terms of another trigonometric function. It is possible as per reciprocal relation of cosecant function with sin function.

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

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

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

Use Sum to Product Trigonometric identity

The numerator of the function represents the sum of sine functions and it can be simplified by applying sum to product transformation identity and it allows us to transform sum of sin functions as a product of trigonometric functions.

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

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

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

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

Simplify expression by negative trigonometric identity

As per negative trigonometric identity, the sine of negative angle can be written as negative of sin of angle.

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

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

Split the expression as multiplying factors

In this expression, part of the expression contains lim sinx/x as x approaches 0 formula. So, split it as two multiplying factors.

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

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

Apply Limit Product Rule

The limit value belongs to both multiplying factors. So, display it to both multiplying factors separately.

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

If $h \,\to\, 0$ then $\dfrac{h}{2} \,\to\, \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \,\to\, 0$. Now, change the limit of second function but no need to change the limit of second multiplying function.

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

Obtain Derivative of Coseant function

The mathematical expression represents differentiation of $operatorname{cosec}{x}$ with respect to $x$. Substitute $h = 0$ in the first multiplying function and apply lim sinx/x as x approaches 0 formula for the second multiplying factor.

$\implies \dfrac{d}{dx} \csc{x}$ $\,=\,$ $-\dfrac{\cos{\Bigg[\dfrac{2x+0}{2}\Bigg]}}{\sin{(x+0)}\sin{x}} \times 1$

$\implies \dfrac{d}{dx} \csc{x}$ $\,=\,$ $-\dfrac{\cos{\Bigg[\dfrac{2x}{2}\Bigg]}}{\sin{x}\sin{x}}$

$\implies \dfrac{d}{dx} \csc{x}$ $\,=\,$ $\require{cancel} -\dfrac{\cos{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}}{\sin{x}\sin{x}}$

$\implies \dfrac{d}{dx} \csc{x}$ $\,=\,$ $-\dfrac{\cos{x}}{\sin{x}\sin{x}}$

Split the trigonometric expression as two multiplying factors to express it in simple form.

$\implies \dfrac{d}{dx} \csc{x}$ $\,=\,$ $-\dfrac{1}{\sin{x}} \times \dfrac{\cos{x}}{\sin{x}}$

According to reciprocal identity of sine function, it can be expressed as cosecant function and the quotient of cosine by sine functions can be expressed as cotangent as quotient identity of cosine and sine functions.

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

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

It is also written in the following way in some countries.

$\implies \dfrac{d}{dx} \, \operatorname{cosec}{x} = -\operatorname{cosec}{x}\cot{x}$

Therefore, it has proved that the derivative of cosecant of angle $x$ with respect to $x$ is equal to the negative of product of cosecant of $x$ and cotangent of $x$.

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