Question

Solve the given expression, $\dfrac{d(cosecx)}{dx}$ .

Answer 1

Hint: We know that, $\text{cosecx=}\dfrac{\text{1}}{\text{sinx}}={{(sinx)}^{-1}}$ . We have the term $\operatorname{sinx}$ in the numerator of the expression $\dfrac{d{{(sinx)}^{-1}}}{dx}$ . Now using chain rule we can write $\dfrac{d{{(sinx)}^{-1}}}{dx}$ as $\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}$ . We know that, $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\sin x}{dx}=\cos x$ . Now, using these formulas we can solve further.

Complete step-by-step answer:
According to the question, we have to solve $\dfrac{d(cosecx)}{dx}$ ……………………(1)
Here, our function to differentiate is $\text{cosecx}$ .
We know the relation between sine function and cosec function. That is, cosec is reciprocal of sine function.
$\sin x=\dfrac{1}{\cos ecx}={{(sinx)}^{-1}}$ …………………(2)
Using equation (1) and equation (2), we get
$\dfrac{d(cosecx)}{dx}=\dfrac{d{{(sinx)}^{-1}}}{dx}$ …………………(3)
In equation (3), we can see that direct differentiation is complex. We need to simplify it in simpler form.
Using chain rule, we can simplify it.
$\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}$ ……………………….(4)
We know the formula,
$\dfrac{d\sin x}{dx}=\cos x$ …………………….(5)
From equation (4), and equation (5), we get
$\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}$
$\text{=}\dfrac{\text{d(sinx}{{\text{)}}^{\text{-1}}}}{\text{dsinx}}\text{ }\\!\\!\times\\!\\!\text{ cosx}$ …………………………(6)
Replacing $\text{sinx}$ by y in the above equation(6), we get
$\dfrac{d{{y}^{-1}}}{dy}\times \cos x$ …………………..(7)
We know the formula that,
$\dfrac{d{{y}^{n}}}{dy}=n{{y}^{n-1}}$
Putting n=-1 in the above equation, we get
$\dfrac{d{{y}^{-1}}}{dy}=-{{y}^{-1-1}}=-{{y}^{-2}}$ …………………….(8)
From equation (7) and equation (8), we get
$\dfrac{d{{y}^{-1}}}{dy}\times \cos x$
$=-{{y}^{-2}}.\cos x$ …………………(9)
We had replaced $\text{sinx}$ by y. So here, we are replacing y by $\text{sinx}$ .
Now, replacing y by $\text{sinx}$ in equation (9), we get
$=-{{y}^{-2}}.\cos x$
$=-{{(sinx)}^{-2}}\cos x$
$=-\dfrac{\cos x}{{{(sinx)}^{2}}}$ …………………..(10)
We know that, $\text{cosecx=}\dfrac{\text{1}}{\text{sinx}}$ and $\dfrac{\cos x}{\sin x}=\cot x$ .
Transforming equation (10), we get
$=-\dfrac{\cos x}{{{(sinx)}^{2}}}$
$=-\dfrac{\cos x}{(sinx)(sinx)}$
$=-cotx.cosecx$
Hence, $\dfrac{d(cosecx)}{dx}=-cotx.cosecx$ .

Note: In chain rule one may think as $\dfrac{d{{(sinx)}^{-1}}}{dx}=\dfrac{d{{(sinx)}^{-1}}}{d\cos x}\times \dfrac{d\cos x}{dx}$ . If we do so then our differentiation will become complex to solve. So, we can’t solve this question by this approach. We can see that the numerator of the expression $\dfrac{d{{(sinx)}^{-1}}}{dx}$ has the term $\operatorname{sinx}$ . So, in chain rule we have to use $\operatorname{sinx}$ in order to make the differentiation easy to solve.