Question

Find the derivative of $\sin \left( {x + 1} \right)$ with respect to $x$ from first principles.

Answer 1

Using the first principle, we can write $\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$. We will use this information to find the derivative of the required function. Also we will use trigonometric results.

Complete step-by-step answer:
In this problem, we have to find the derivative of $\sin \left( {x + 1} \right)$ by using first principle. For this, let us consider $y = \sin \left( {x + 1} \right)$ or $f\left( x \right) = \sin \left( {x + 1} \right)$. Let us replace $x$ by $x + h$. Therefore, we get $f\left( {x + h} \right) = \sin \left( {x + h + 1} \right)$.
Using the first principle, we can write $\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cdots \cdots \left( 1 \right)$. Now we will substitute $f\left( {x + h} \right)$ and $f\left( x \right)$ in equation $\left( 1 \right)$ and evaluate the limit. Therefore, we get
$\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h + 1} \right) - \sin \left( {x + 1} \right)}}{h} \cdots \cdots \left( 2 \right) \\\$
Now on the RHS of equation $\left( 2 \right)$, we will use the trigonometric identity which is given by
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$. Therefore, we get
$\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \left[ {\dfrac{{\left( {x + h + 1} \right) + \left( {x + 1} \right)}}{2}} \right]\sin \left[ {\dfrac{{\left( {x + h + 1} \right) - \left( {x + 1} \right)}}{2}} \right]}}{h} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \left( {\dfrac{{2x + h + 2}}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{h} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \left[ {\cos \left( {\dfrac{{2x + h + 2}}{2}} \right)\dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}} \right] \cdots \cdots \left( 3 \right) \\\$
Now we will use the multiplication rule for limits. That is, the product of the limits is the same as the limit of the product of two functions. Therefore, from equation $\left( 3 \right)$, we will get
$\dfrac{{dy}}{{dx}} = \left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\dfrac{{2x + h + 2}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}} \right] \cdots \cdots \left( 4 \right)$
Now we are going to use the result $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ on RHS of equation $\left( 4 \right)$. Therefore, we get
$\dfrac{{dy}}{{dx}} = \left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\dfrac{{2x + h + 2}}{2}} \right)} \right]\left( 1 \right)\quad \left[ {\because h \to 0 \Rightarrow \dfrac{h}{2} \to 0} \right] \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \cos \left( {\dfrac{{2x + h + 2}}{2}} \right) \cdots \cdots \left( 5 \right) \\\$
Let us put $h = 0$ on the RHS of equation $\left( 5 \right)$ to find the limit. Therefore, we get
$\dfrac{{dy}}{{dx}} = \cos \left( {\dfrac{{2x + 0 + 2}}{2}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left[ {\dfrac{{2\left( {x + 1} \right)}}{2}} \right] \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {x + 1} \right) \\\$
Therefore, the derivative of $\sin \left( {x + 1} \right)$ with respect to $x$ is $\cos \left( {x + 1} \right)$.

Note: In this problem, it is mentioned that we have to find the derivative by using first principle. Also we can find the derivative of $\sin \left( {x + 1} \right)$ by using the basic formula of derivative and chain rule.