Question

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

Answer 1

Hint: Here we will first describe the method of the first principle of derivative and then further we may apply to find the derivative of the given function ${{\cos }^{2}}x$.

Complete step-by-step answer:

The first principle of derivative refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method.
We know that the gradient of the tangent to a curve with equation y = f(x) at x = a can be determined by the formula:
$Gradient=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$
We can use this formula to determine an expression that describes the gradient of the graph ( or the gradient of the tangent to the graph ) at any point on the graph.
This expression is called the derivative and the process of determining the derivative of a function is called differentiation.
Now, we may apply this formula to find the derivative of ${{\cos }^{2}}x$.
Let $f\left( x \right)={{\cos }^{2}}x$
$\begin{align}
  & f'\left( x \right)=\underset{h\to 0}{\mathop{lt}}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{\left( x+h \right)-h} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underset{h\to 0}{\mathop{lt}}\,\dfrac{{{\cos }^{2}}\left( x+h \right)-{{\cos }^{2}}x}{h} \\
\end{align}$
Since, we know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.
Therefore,
$\begin{align}
  & f'\left( x \right)=\underset{h\to 0}{\mathop{lt}}\,\dfrac{1-{{\sin }^{2}}\left( x+h \right)-\left( 1-{{\sin }^{2x}} \right)}{h} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underset{h\to 0}{\mathop{lt}}\,\dfrac{{{\sin }^{2}}x-{{\sin }^{2}}\left( x+h \right)}{h} \\
\end{align}$
Since, we have a trigonometric formula:
${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A-B \right)\sin \left( A+B \right)$
So, on using this we get to get the derivative of the given function, we get:
$\begin{align}
  & f'\left( x \right)=\underset{h\to 0}{\mathop{lt}}\,\dfrac{\sin \left( x+x+h \right)\sin \left\{ x-\left( x+h \right) \right\}}{h} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underset{h\to 0}{\mathop{lt}}\,\dfrac{\sin \left( 2x+h \right)\sin \left( x-x-h \right)}{h} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\underset{h\to 0}{\mathop{lt}}\,\sin \left( 2x+h \right)\times \left( -\underset{h\to 0}{\mathop{lt}}\,\dfrac{\sinh }{h} \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin 2x\times \left( -1 \right) \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\sin 2x \\
\end{align}$
Hence, the derivative of ${{\cos }^{2}}x$ is $-\sin 2x$.

Note: Students should remember certain trigonometric formulas while solving the problem and should keep in mind that the value of h always tends to zero. So, it can be neglected when it is being added to any other number.