Question

Find the differential coefficient of \[\sin x\] by first principal.

Answer 1

Hint: In this problem, we need to use the first principle to obtain the derivative of the given expression. Next, use the trigonometric identities to solve the limit used in first principle. In this problem, the differential coefficient of \[\sin x\] is a measure of rate of change with respect to\[x\].

Complete step by step answer:
The derivative of a function is a measure of the rate of change. The derivative of a function \[y = f\left( x \right)\], using the first principle is shown below.
\[\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]
Now, substitute \[\sin x\] for \[f\left( x \right)\] in the above formula, to obtain the differential coefficient of \[\sin x\].
\[
  \,\,\,\,\,\,\,\,\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h} \right) - \sin \left( x \right)}}{h} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin x\cosh + \cos x\sinh - \sin x}}{h}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin x\left( {\cosh - 1} \right) + \cos x\sinh }}{h} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin x\left( {\cosh - 1} \right)}}{h} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\sinh }}{h} \\
\]
Further, simplify the above expression.
\[
  \,\,\,\,\,\,\,\dfrac{{dy}}{{dx}} = \sin x\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\cosh - 1} \right)}}{h} + \cos x\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \sin x\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - \dfrac{{{h^2}}}{{2!}} + \dfrac{{{h^4}}}{{4!}} - \ldots \ldots - 1} \right)}}{h} + \cos x\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \ldots \ldots } \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \sin x\mathop {\lim }\limits_{h \to 0} \dfrac{{\left( { - \dfrac{{{h^2}}}{{2!}} + \dfrac{{{h^4}}}{{4!}} - \ldots \ldots } \right)}}{h} + \cos x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \sin x\mathop {\lim }\limits_{h \to 0} \left( { - \dfrac{h}{{2!}} + \dfrac{{{h^3}}}{{4!}} - \ldots \ldots } \right) + \cos x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \sin x\left( 0 \right) + \cos x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \cos x \\
\]

Thus, the differential coefficient of \[\sin x\] is \[\cos x\].

Note: If \[f\] be a function of time, then the derivative of \[f\] with respect to time represents the rate of change of \[f\] with respect to time.
The expansion of the \[\cos x\] using Taylor series is shown below.
\[\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \ldots \ldots\]