Question

Find the derivative of ${\csc ^2}x$, by using the first principle of derivatives.

Answer 1

To solve this problem,consider Let $f(x) = {\csc ^2}x$. Find $f(x + h)$ and hence find $f'(x)$using the formula $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$.

Complete step by step solution:
We are given a trigonometric function ${\csc ^2}x$.
We need to find its derivative by using the first principle of derivatives.
Now, let $f(x) = {\csc ^2}x$
The formula for finding the derivatives using the first principle is as follows:
$g'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{g(x + h) - g(x)}}{h}$
where$g'(x)$denotes the derivative of the function$g(x)$and h is a very small value such that g is defined for both$x$and$x{\text{ }} + {\text{ }}h$
Therefore,
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
Now, $f(x) = {\csc ^2}x \Rightarrow f(x + h) = {\csc ^2}(x + h)$
Then
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\csc }^2}(x + h) - {{\csc }^2}x}}{h}$
Using the identity $\csc \theta = \dfrac{1}{{\sin \theta }}$, we get

$$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{{{\sin }^2}(x + h)}} - \dfrac{1}{{{{\sin }^2}x}}}}{h} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{{{\sin }^2}x - {{\sin }^2}(x + h)}}{{{{\sin }^2}(x + h) \times {{\sin }^2}x}}}}{h} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times ({\sin ^2}x - {\sin ^2}(x + h)) \times \dfrac{1}{{{{\sin }^2}(x + h) \times {{\sin }^2}x}} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times ({\sin ^2}x - {\sin ^2}(x + h)) \times \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{{{\sin }^2}(x + h) \times {{\sin }^2}x}}...........(1) \\\$$

Now,

$$\mathop {\lim }\limits_{h \to 0} \dfrac{1}{{{{\sin }^2}(x + h) \times {{\sin }^2}x}} \\\ = \dfrac{1}{{{{\sin }^2}(x + 0) \times {{\sin }^2}x}} \\\ = \dfrac{1}{{{{\sin }^2}x \times {{\sin }^2}x}} \\\ = \dfrac{1}{{{{\sin }^4}x}} \\\ = {\csc ^4}x.........(2) \\\$$

And using the identity ${a^2} - {b^2} = (a - b)(a + b)$, we have

$$\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times ({\sin ^2}x - {\sin ^2}(x + h)) \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times (\sin x - \sin (x + h)) \times (\sin x - \sin (x + h)).......(3) \\\$$

We know that
$\sin A \pm \sin B = 2\sin \left( {\dfrac{{A \pm B}}{2}} \right)cos\left( {\dfrac{{A \mp B}}{2}} \right)$
Here, in (3), we have$A = x,B = x + h$
We use this identity to simplify (3)

$$\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times ({\sin ^2}x - {\sin ^2}(x + h)) \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times (2\sin \dfrac{{(x - (x + h))}}{2})\cos (\dfrac{{x + (x + h)}}{2}) \times 2\sin \dfrac{{(x + (x + h))}}{2})\cos (\dfrac{{x - (x + h)}}{2}) \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times 2\sin (\dfrac{{ - h}}{2})\cos (x + \dfrac{h}{2}) \times 2\sin (x + \dfrac{h}{2})\cos (\dfrac{{ - h}}{2}).....(4) \\\$$

We will use the following facts in (4).

$$\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin \theta }}{\theta } = 1 \\\ \cos 0 = 1 \\\ h \to 0 \Rightarrow \dfrac{{ - h}}{2} \to 0 \\\$$ $$\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times ({\sin ^2}x - {\sin ^2}(x + h)) \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} \times 2\sin (\dfrac{{ - h}}{2}) \times \mathop {\lim }\limits_{h \to 0} \cos (x + \dfrac{h}{2}) \times 2\mathop {\lim }\limits_{h \to 0} \sin (x + \dfrac{h}{2}) \times \mathop {\lim }\limits_{h \to 0} \cos (\dfrac{{ - h}}{2}) \\\ = \mathop { - \lim }\limits_{h \to 0} \dfrac{{\sin (\dfrac{{ - h}}{2})}}{{\dfrac{{ - h}}{2}}} \times \mathop {\lim }\limits_{h \to 0} \cos (x + \dfrac{h}{2}) \times 2\mathop {\lim }\limits_{h \to 0} \sin (x + \dfrac{h}{2}) \times \mathop {\lim }\limits_{h \to 0} \cos (\dfrac{{ - h}}{2}) \\\ = - 1 \times \cos (x + 0) \times 2 \times \sin (x + 0) \times \cos 0 \\\ = - 2\sin x\cos x.........(5) \\\$$

Combining (1), (2), and (5), we get

$$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\csc }^2}(x + h) - {{\csc }^2}x}}{h} \\\ = - 2\sin x\cos x{\csc ^4}x \\\ = - 2\sin x\cos x\dfrac{1}{{{{\sin }^4}x}} \\\ = - 2{\csc ^2}x\cot x \\\$$

Hence the derivative of ${\csc ^2}x$is$- 2{\csc ^2}x\cot x$.

Note: One must remember that$\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (\dfrac{1}{h})}}{{\dfrac{1}{h}}} \ne 0$. This is because $h \to 0$,$\dfrac{1}{h} \to \infty$. Therefore, the formula$\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin \theta }}{\theta } = 1$does not work here.