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**Hint:**We recall the first principle of derivative. We assume a small change in $x$ as $\delta x$ and its corresponding change in $y=f\left( x \right)$ as $\delta y$. We find the average rate of change as $\dfrac{\delta y}{\delta x}=\dfrac{f\left( x+\delta x \right)-f\left( x \right)}{\delta x}$ . We take limit $\delta x \to 0$ to find the instantaneous rate of change as derivative of $f\left( x \right)$.\[\]

**Complete step-by-step solution:**

We are given the function $f\left( x \right)={{\cos }^{2}}x$ in the question. Let us have$y={{\cos }^{2}}x$. Let $\delta x$ be a very small change in $x$ and the corresponding change in $y$ be $\delta y$. So we have;

\[y+\delta y={{\cos }^{2}}\left( x+\delta x \right)\]

We subtract $y$ both sides of the above equation to have;

\[\begin{align}

& \Rightarrow y+\delta y-y={{\cos }^{2}}\left( x+\delta x \right)-y \\

& \Rightarrow y+\delta y-y={{\cos }^{2}}\left( x+\delta x \right)-{{\cos }^{2}}x \\

& \Rightarrow \delta y={{\cos }^{2}}\left( x+\delta x \right)-{{\cos }^{2}}x \\

\end{align}\]

We divide $\delta x$ both sides of the above step to have;

\[\Rightarrow \dfrac{\delta y}{\delta x}=\dfrac{{{\cos }^{2}}\left( x+\delta x \right)-{{\cos }^{2}}x}{\delta x}\]

We take limit $\delta x \to 0$ both sides of the above step to have;

\[\Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=\displaystyle \lim_{\delta x \to 0}\dfrac{{{\cos }^{2}}\left( x+\delta x \right)-{{\cos }^{2}}x}{\delta x}\]

We use the trigonometric identity ${{\cos }^{2}}B-{{\cos }^{2}}A=\sin \left( A+B \right)\sin \left( A-B \right)$ for $A=x,B=x+\delta x$ in the above step to have;

\[\begin{align}

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=\displaystyle \lim_{\delta x \to 0}\dfrac{\sin \left( x+\delta x+x \right)\sin \left( x-\delta x-x \right)}{\delta x} \\

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=\displaystyle \lim_{\delta x \to 0}\dfrac{\sin \left( 2x+\delta x \right)\sin \left( -\delta x \right)}{\delta x} \\

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-\displaystyle \lim_{\delta x \to 0}\dfrac{\sin \left( 2x+\delta x \right)\sin \left( \delta x \right)}{\delta x} \\

\end{align}\]

We use law of product if limits in the right hand side of the above step to have;

\[\Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-\displaystyle \lim_{\delta x \to 0}\sin \left( 2x+\delta x \right)\cdot \displaystyle \lim_{\delta x \to 0}\dfrac{\sin \left( \delta x \right)}{\delta x}\]

We use the standard limit $\displaystyle \lim_{x \to o}\dfrac{\sin x}{x}=1$ for $x=\delta x$ in the right hand side of the above step to have;

\[\begin{align}

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-\displaystyle \lim_{\delta x \to 0}\sin \left( 2x+\delta x \right)\cdot 1 \\

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-\displaystyle \lim_{\delta x \to 0}\sin \left( 2x+\delta x \right) \\

& \Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-\sin 2x \\

\end{align}\]

We use the double angle formula $\sin 2\theta =2\sin \theta \cos \theta $ for $\theta =x$ in the right hand side of the above step to have;

\[\Rightarrow \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=-2\sin x\cos \]

We know from first principle of derivative that $\displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=\dfrac{dy}{dx}$. So we have

\[\begin{align}

& \Rightarrow \dfrac{dy}{dx}=-2\sin x\cos x \\

& \Rightarrow \dfrac{d}{dx}\left( {{\cos }^{2}}x \right)=-2\sin x\cos x \\

\end{align}\]

**Note:**We can use chain rule to directly find the derivative of ${{\cos }^{2}}x$. If composite function is defined as $y=u\left( x \right),u=f\left( x \right)$ then the chain rule is given as $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}$. We can also use the first principle for derivative with a very small change $h$ as $\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( h \right)}{h}$. The derivative of the function at particular points geometrically gives the slope of the tangent to the curve of the function. The first principle is also known as the delta method.

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