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Question: Show that the function \(g\left( x \right)=\cos x\) is differentiable at any \(a\in \mathbb{R}\) and...

Show that the function g(x)=cos⁡xg\left( x \right)=\cos x is differentiable at any a∈Ra\in \mathbb{R} and g′(a)=−sin⁡a.g'\left( a \right)=-\sin a. In general, g′(x)=−sin⁡xg'\left( x \right)=-\sin x.

Explanation

Solution

Hint: To check the differentiability of a function, we will find the derivative of the function. To find the derivative of any function f(x)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(x)=sin⁡xg\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(x)g\left( x \right), we will differentiate g(x)g\left( x \right) using the first principle of derivative.
According to first principle, we can find derivative g′(x)g'\left( x \right) of the functiong(x)g\left( x \right) using the formula;
g′(x)=lim⁡h→0 g(x+h)−g(x)h.........(I)g'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( x+h \right)-g\left( x \right)}{h}.........\left( I \right)
Substituting g(x)=cos⁡xg\left( x \right)=\cos x and g(x+h)=cos⁡(x+h)g\left( x+h \right)=\cos \left( x+h \right) in the formula (I)\left( I \right), we get;
g′(x)=lim⁡h→0 cos⁡(x+h)−cos⁡xh.........(II)g'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+h \right)-\cos x}{h}.........\left( II \right)
In trigonometry, we have a formula; cos⁡(x+h)=cos⁡xcos⁡h−sin⁡xsin⁡h\cos \left( x+h \right)=\cos x\cos h-\sin x\sin h
Substituting this formula i.e. cos⁡(x+h)=cos⁡xcos⁡h−sin⁡xsin⁡h\cos \left( x+h \right)=\cos x\cos h-\sin x\sin h in (II)\left( II \right), we get;

& \Rightarrow g'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h} \\\ & \Rightarrow g'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos x\cos h-\cos x}{h}-\dfrac{\sin x\sin h}{h} \\\ \end{aligned}$$ Since the limit can be distributed over subtraction of the two functions, we can write; $$\Rightarrow g'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos x\left( \cos h-1 \right)}{h}-\underset{h\to 0}{\mathop{\lim }}\,\sin x\dfrac{\sin h}{h}$$ Since the limit is applied on $h$, we can take the functions of$x$out of the individual limits. $\Rightarrow g'\left( x \right)=\cos x\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos h-1}{h}-\sin x\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin h}{h}........\left( III \right)$ We have two formulas of limit, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos h-1}{h}=0$ and $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin h}{h}=1$ . Substituting $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos h-1}{h}=0$ and $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin h}{h}=1$in equation $\left( III \right)$, we get; $\begin{aligned} & \Rightarrow g'\left( x \right)=\cos x\left( 0 \right)-\sin x\left( 1 \right) \\\ & \Rightarrow g'\left( x \right)=0-\sin x \\\ & \Rightarrow g'\left( x \right)=-\sin x \\\ \end{aligned}$ It will be easier to check the differentiability of $g\left( x \right)$ if we draw the graph of $g'\left( x \right)=\left( -\sin x \right)$. Plotting the graph of $g'\left( x \right)=\left( -\sin x \right)$; ![](https://www.vedantu.com/question-sets/0b6a8e1b-f599-4e12-b44e-ef8f5dcc7c8e3169566565307588118.png) Since the graph of $g'\left( x \right)$ is continuous $\forall x\in \mathbb{R}.$, we can say $g\left( x \right)$ is differentiable$\forall x\in \mathbb{R}.$. Also, $g'\left( x \right)=-\sin x$. Hence, substituting $x=a$ in $g'\left( x \right)=-\sin x$, we obtain; $\text{g}'\left( a \right)=-\sin a$ 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 $g'\left( x \right)$ which may make this question a time taking one.