Question

Find the derivative of \[sin\text{ }x\] with respect to x from first principles.

Answer 1

Hint: The derivative using the first principal is given by the formula \[\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]\], where f(x) is the function to be differentiation with respect to x, and h is tending to zero.
Complete step-by-step answer:
In the question, we have to find the derivative of \[sin\text{ }x\] with respect to x from first principles.
So, for that we will use the formula \[\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]\], here the function to be differentiated is \[f\left( x \right)=\text{ }sin\text{ }x\]. Now as per the formula, we have the required derivative as:
\[\begin{align}
  & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right] \\
 & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\
\end{align}\]
Here we can see that the limit is of the form \[\dfrac{0}{0}\], as we have \[\underset{h\to \;0}{\mathop{\lim }}\,\,\,\sin \left( x+h \right)=\sin \left( x \right)\]
So \[\underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right)=\dfrac{0}{0}\]
Hence, here we will use the L-Hopitals rule, where we differentiate the numerator and the denominator separately, as follows:
\[\begin{align}
  & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\
 & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{d}{dh}\left( \sin \left( x+h \right)-\sin \left( x \right) \right)}{\dfrac{d}{dh}\left( h \right)} \right) \\
 & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\left( \cos \left( x+h \right)-0 \right)}{\left( 1 \right)} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \left( \dfrac{d\left( \sin \left( x+h \right) \right)}{dh}=\cos \left( x+h \right),\,\,\,\dfrac{d\left( \sin \left( x \right) \right)}{dh}=0 \right) \\
 & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \cos \left( x+h \right) \right) \\
 & \Rightarrow \cos \left( x \right) \\
\end{align}\]
So final, we can say that
\[\begin{align}
  & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right] \\
 & \Rightarrow \underset{h\to \;0}{\mathop{\lim }}\,\left( \dfrac{\sin \left( x+h \right)-\sin \left( x \right)}{h} \right) \\
 & \Rightarrow \cos \left( x \right) \\
\end{align}\]
Hence the derivative of \[\sin \left( x \right)\] is \[\cos \left( x \right)\].

Note: The limit of the expression is to be found carefully, and when applying the L-Hopitals’ rule, we will separately differentiate the numerator and the denominator and will not use the quotient rule of differentiation.