Question

Find the derivative of the following function from the first principle: $\sin (x+1)$ .

Answer 1

Hint: The given problem is related to differentiation. According to the first principle, the derivative of a function $f\left( x \right)$ at $x=a$ is given as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$ .

Complete Step-by-step answer:
Before proceeding with the problem, let’s understand the concept of differentiation. The derivative of a function represents the slope of the graph of the function. The derivative of the function $f\left( x \right)$ is denoted by $f'\left( x \right)$ and is defined by the first principle as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . The steps to determine the derivative of a function by the first principle are :
Step 1: Write down the formula for finding the derivative by the first principle.
Step 2: Determine the value of $f\left( x+h \right)$ .
Step 3: Substitute it into the formula.
Step 4: Find the limit.
Now, the given function is $\sin \left( x+1 \right)$ .
Following the steps stated above, first we will write the formula for finding the derivative of the function by the first principle.
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ .
Now, we will find the value of f(x+ h) by replacing x by (x + h). The function is given as $f\left( x \right)=\sin \left( x+1 \right)$ . So, the value of $f\left( x+h \right)$ is given as $\sin \left( x+1+h \right)$ , which can be written as $\sin \left( \left( x+1 \right)+h \right)$ . We know, $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ . So, $\sin \left( \left( x+1 \right)+h \right)=\sin \left( x+1 \right)\cos \left( h \right)+\cos \left( x+1 \right)\sin \left( h \right)$ .
On substituting the value of f(x + h) in the expression for derivative of a function, we get:$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+1 \right)\cos \left( h \right)+\cos \left( x+1 \right)\sin \left( h \right)-\sin \left( x+1 \right)}{h}$
Now, we will evaluate the limit. The expression that we have is:$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+1 \right)\cos \left( h \right)+\cos \left( x+1 \right)\sin \left( h \right)-\sin \left( x+1 \right)}{h}$
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+1 \right)\left( \cos \left( h \right)-1 \right)+\cos \left( x+1 \right)\sin \left( h \right)}{h}$
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+1 \right)\left( \cos \left( h \right)-1 \right)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+1 \right)\sin \left( h \right)}{h}$
Now, we know $\sin \left( x+1 \right)$ is a function of $x$ and is independent of $h$ . So, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x+1 \right)\left( \cos \left( h \right)-1 \right)}{h}=\sin \left( x+1 \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \cos \left( h \right)-1 \right)}{h}$. Now, we know, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \cos \left( h \right)-1 \right)}{h}=0$ . So, $f'\left( x \right)=\sin \left( x+1 \right)\times 0+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+1 \right)\sin \left( h \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+1 \right)\sin \left( h \right)}{h}$. Now, we know, $\cos \left( x+1 \right)$ is a function of $x$ and hence, it is independent of $h$ . So, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( x+1 \right)\sin \left( h \right)}{h}$ can be written as $f'\left( x \right)=\cos \left( x+1 \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}$. Now, we know, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}=1$ . So, $f'\left( x \right)=\cos \left( x+1 \right)$ .
Hence, the derivative of $\sin (x+1)$ is $\cos \left( x+1 \right)$ .

Note: While determining the derivative of the function, make sure to use the expansions and formulae of the limit properly. Some students write the expansion of $\sin \left( a+b \right)$ as $\sin a\cos b-\cos a\sin b$ instead of $\sin a\cos b+\cos a\sin b$ . Also, some students write the value of the limit $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}=0$ instead of $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}=1$. Such mistakes should be avoided, as they can result in getting wrong answers.