Question

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

Answer 1

Hint: 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.