Question

Show that the function $g\left(x\right)=\cos x$ is differentiable at any $a\in \mathbb{R}$ and $g^{\prime }\left(a\right)=-\sin a.$ In general, $g^{\prime }\left(x\right)=-\sin x$.

Answer 1

Hint: To check the differentiability of a function, we will find the derivative of the function. To find the derivative of any function $f\left(x\right)$, we will use the first principle of derivative.

Complete step-by-step answer:
 In this question, we are given a function; $$g\left(x\right)=\sin x$$.
Since we have to check for the differentiability of this function, we have to first find its derivative. To find the derivative of the function $g\left(x\right)$, we will differentiate $g\left(x\right)$ using the first principle of derivative.
According to first principle, we can find derivative $g^{\prime }\left(x\right)$ of the function$g\left(x\right)$ using the formula;
$g^{\prime }\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{g(x+h)-g\left(x\right)}{h}}.........\left(I\right)$
Substituting $g\left(x\right)=\cos x$ and $g(x+h)=\cos (x+h)$ in the formula $\left(I\right)$, we get;
$$g^{\prime }\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\cos (x+h)-\cos x}{h}}.........\left(II\right)$$
In trigonometry, we have a formula; $\cos (x+h)=\cos x\cos h-\sin x\sin h$
Substituting this formula i.e. $\cos (x+h)=\cos x\cos h-\sin x\sin h$ in $\left(II\right)$, we get;
$$\begin{array}{rl}& \Rightarrow g^{\prime }\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\cos x\cos h-\sin x\sin h-\cos x}{h}}\\ & \Rightarrow g^{\prime }\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\cos x\cos h-\cos x}{h}}-{\displaystyle \frac{\sin x\sin h}{h}}\end{array}$$
Since the limit can be distributed over subtraction of the two functions, we can write;
$$\Rightarrow g^{\prime }\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\cos x(\cos h-1)}{h}}-\underset{h\to 0}{lim}\sin x{\displaystyle \frac{\sin h}{h}}$$
Since the limit is applied on $h$, we can take the functions of$x$out of the individual limits.
$\Rightarrow g^{\prime }\left(x\right)=\cos x\underset{h\to 0}{lim}{\displaystyle \frac{\cos h-1}{h}}-\sin x\underset{h\to 0}{lim}{\displaystyle \frac{\sin h}{h}}........\left(III\right)$
We have two formulas of limit, $\underset{h\to 0}{lim}{\displaystyle \frac{\cos h-1}{h}}=0$ and $\underset{h\to 0}{lim}{\displaystyle \frac{\sin h}{h}}=1$ .
Substituting $\underset{h\to 0}{lim}{\displaystyle \frac{\cos h-1}{h}}=0$ and $\underset{h\to 0}{lim}{\displaystyle \frac{\sin h}{h}}=1$in equation $\left(III\right)$, we get;
$\begin{array}{rl}& \Rightarrow g^{\prime }\left(x\right)=\cos x\left(0\right)-\sin x\left(1\right)\\ & \Rightarrow g^{\prime }\left(x\right)=0-\sin x\\ & \Rightarrow g^{\prime }\left(x\right)=-\sin x\end{array}$
It will be easier to check the differentiability of $g\left(x\right)$ if we draw the graph of $g^{\prime }\left(x\right)=(-\sin x)$.
Plotting the graph of $g^{\prime }\left(x\right)=(-\sin x)$;

Since the graph of $g^{\prime }\left(x\right)$ is continuous $\mathrm{\forall }x\in \mathbb{R}.$, we can say $g\left(x\right)$ is differentiable$\mathrm{\forall }x\in \mathbb{R}.$.
Also, $g^{\prime }\left(x\right)=-\sin x$. Hence, substituting $x=a$ in $g^{\prime }\left(x\right)=-\sin x$, we obtain;
${\text{g}}^{\prime }\left(a\right)=-\sin a$

Note: The question can be done directly if one has remembered that the derivatives of $g\left(x\right)=\cos x$ instead of applying the first principle and to finding the value of $g^{\prime }\left(x\right)$ which may make this question a time taking one.