Question

Find the derivative of $\sin x$ with respect to x from first principles.

Answer 1

In this question, we have been asked to find the derivative of a basic trigonometric equation but it is specifically mentioned that it is to be found using the first principle method. Start by assuming our given function to be equal to $f\left( x \right)$. Then, find $f\left( {x + h} \right)$. Put this in the first principle method. Next, use some trigonometric formula to simplify the sin and cos. Once the equation has been completely simplified, apply limits.

Formula used: 1) $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
2) $\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$

Complete step-by-step solution:
We have asked to find the derivative of $\sin x$ using first principle method. I will try to explain its solution step by step.
If we assume our function to be $f\left( x \right)$, then its derivative using first principle method is given by - $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$.
Let us move towards the question.
Step 1: Let $f\left( x \right) = \sin x$.
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
Step 2: This step includes finding $f\left( {x + h} \right)$.
$\Rightarrow f\left( {x + h} \right) = \sin \left( {x + h} \right)$
Step 3: In this step, we will put the value of $f\left( {x + h} \right)$ in the formula.
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h} \right) - \sin \left( x \right)}}{h}$
Step 4: Simplify.
Now, we will simplify by putting certain formulas of sin.
We will use the formula $\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$ and expand .
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \left( {\dfrac{{x + h + x}}{2}} \right)\sin \left( {\dfrac{{x + h - x}}{2}} \right)}}{h}$
Dividing numerator and denominator by 2.
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\dfrac{{x + h + x}}{2}} \right)\sin \left( {\dfrac{{x + h - x}}{2}} \right)}}{{\dfrac{h}{2}}}$
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\dfrac{{2x + h}}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}$
Separating the numerator into two parts,
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \left( {\dfrac{{2x + h}}{2}} \right) \times \dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}$
We know that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$. Using this in the above equation,
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \left( {\dfrac{{2x + h}}{2}} \right)$
Step 4: This step includes applying limits $h = 0$.
$\Rightarrow f'\left( x \right) = \cos \left( {\dfrac{{2x}}{2}} \right)$
$\Rightarrow f'\left( x \right) = \cos x$

Therefore, the derivative of $\sin x$ with respect to x using the first principle method is $\cos x$.

Note: We have to mind that, the first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the anti-derivatives (also called indefinite integral), say $F$, of some function $f$ may be obtained as the integral of $f$ with a variable bound of integration.
Conversely, the second part of the theorem, sometimes called the second fundamental theorem of calculus, states that the integral of a function $f$ over some interval can be computed by using anyone, say $F$ of its infinitely many anti-derivatives.
The answer will remain the same whether you find it using the first principle method or directly. Therefore, find the derivative of the function directly and verify your answer. Do not open $\sin \left( {x + h} \right)$ using the identity $\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y$ as it won’t do any good. Always try to use the identities that will help in simplifying the question.