Question

How do you find the derivative of $(\cos x)$using the limit definition ?

Answer 1

Hint: In order to find the first derivative of the above expression using the limit definition , we need to solve it step by step . Also we need to know some important concepts of the terms used in the question which will help us in solving the question , So, the two terms Derivatives and Limits are used and we should have knowledge about the same before solving the question . So , Limit is defined as a value that a function approaches as the input, and it produces some value . We will be using some limited formula and properties required to solve the question . And the derivative refers to the instantaneous rate of change of a quantity with respect to the other , The derivative of a function is represented in the below-given formula. $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$.

Complete step-by-step answer:
 Given a function $(\cos x)$ , we know the formula of the derivative $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
So, $f(x) = \cos x$
Applying limits and derivative to the function , we get –
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x + h) - \cos (x)}}{h}$
Now we will use the trigonometry formula $\cos (x + h) = \cos (x)\cos (h) - \sin (x)\sin (h)$ to rewrite the derivative of $(\cos x)$ as –
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)\cos (h) - \sin (x)\sin (h) - \cos (x)}}{h}$

Rewrite as follows –
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)\cos (h) - \sin (x)\sin (h) - \cos (x)}}{h}$
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)(\cos (h) - 1) - \sin (x)\sin (h) - \cos (x)}}{h}$
Use the theorem on limits that states: the limit of a difference of two functions is equal to the difference of the limits, to rewrite as follows –
\[f'(x)f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)(\cos (h) - 1)}}{h} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (x)\sin (h)}}{h}\]----------------equation 1
Now there are some fundamental trigonometric limits :
$\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin \theta }}{\theta } = 1$ and $\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\cos \theta - 1}}{\theta } = 0$
Applying these limits in equation 1 we get –

\[
  f'(x) = \cos (x)\left( {\mathop {\lim }\limits_{h \to 0} \dfrac{{(\cos (h) - 1)}}{h}} \right) - \sin (x)\left( {\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (h)}}{h}} \right) \\
  f'(x) = \cos (x)(0) - \sin (x)(1) \\
  f'(x) = 0 - \sin (x) \\
  f'(x) = - \sin (x) \\
 \]
Therefore , the derivative of $(\cos x)$using the limit definition is \[f'(x) = - \sin (x)\]
$f(x) = \cos x$
\[\cos '(x) = - \sin (x)\]
So, the correct answer is “$ - \sin (x) $”.

Note: Always try to understand the mathematical statement carefully and keep things distinct .
Remember the properties and apply appropriately .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .
Avoid jumping the steps as it can create an error. To solve the problem, ideas about limits are utmost important.