Find derivative of 1/sinx (1 by sinx)

The derivative of 1/sinx is equal to -cosecx cotx. In this post, we will learn how to differentiate 1 by sinx with respect to x.

What is the Derivative of 1/sinx?

Answer: The derivative of 1/sinx with respect to x is denoted by d/dx(1/sinx) and it is equal to -cosec(x)cot(x). That is,

$\dfrac{d}{dx}(\dfrac{1}{\sin x})$ = $-\text{cosec} x \cot x$.

Proof:

Note that 1/sinx is a quotient function. So to find its derivative, we can apply the quotient rule of derivatives.

Lets recall the quotient rule of derivatives:

$\dfrac{d}{dx} \left(\dfrac{f}{g} \right)$ = $\dfrac{g \frac{df}{dx} – f \frac{dg}{dx}}{g^2}$

In this formula, we put f=1 and g=sinx.

Noting the derivative of a constant is zero and the derivative of sinx is cosx, we obtain that

$\dfrac{d}{dx}(\dfrac{1}{\sin x})$

= $\dfrac{\sin x \frac{d}{dx}(1) – 1 \frac{d}{dx}(\sin x)}{\sin^2 x}$

= $\dfrac{\sin x \cdot 0 – 1 \cdot \cos x}{\sin^2 x}$

= $-\dfrac{\cos x}{\sin^2 x}$

= $- \dfrac{1}{\sin x} \cdot \dfrac{\cos x}{\sin x}$

= $-\text{cosec} x \cot x$.

So the derivative of 1/sinx is -cosec(x)cot(x) and this is obtained by applying the quotient rule of derivatives.

Video Solution on Derivative of 1/sinx:

Remark:

As the reciprocal of sinx is cosecx, we have 1/sinx = cosecx. Also, we know that the derivative of cosecx is -cosecx cotx. Thus, the derivative of 1/sinx is same as the derivative of cosecx; and we have that

d/dx(1/sinx) = -cosecx cotx.

Question: Find the derivative of 1/sin2x.

From the above, we know that the derivative of 1/sinx is -cosecx cotx. So by the chain rule of derivatives,

d/dx(1/sin2x) = d/dz(1/sinz) ⋅ dz/dx where z=2x.

= -cosecz cotz ⋅ 2

= -2cosec(2x) cot(2x) as z=2x.

So the derivative of 1 over sin2x is equal to -2cosec(2x) cot(2x).