Question

Differentiate \[\cos \left( {{x^2} + 1} \right)\] using the first principle of derivative.

Answer 1

Hint: In the above question, we are a trigonometric function that is \[\cos \left( {{x^2} + 1} \right)\] . we have to differentiate the given trigonometric function using the first principle of derivative. The first principle of differentiation helps us evaluate the derivative of a function using limits. According to the first principle of derivative, the derivative of a function \[f\left( x \right)\] , that is \[f'\left( x \right)\] , is given in the form of limits by the formula:
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]
In order to approach our solution, we have to use the above formula.

Complete step-by-step answer:
The trigonometric function is \[\cos \left( {{x^2} + 1} \right)\] .
Let,
\[ \Rightarrow f\left( x \right) = \cos \left( {{x^2} + 1} \right)\]
We have to find the derivative of \[f\left( x \right)\] i.e. \[f'\left( x \right)\] .
According to the first principle of derivative, we have the equation
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]
Since \[f\left( x \right) = \cos \left( {{x^2} + 1} \right)\] , hence \[f\left( {x + h} \right) = \cos \left( {{{\left( {x + h} \right)}^2} + 1} \right)\]
Therefore we can write the above equation as,
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {{{\left( {x + h} \right)}^2} + 1} \right) - \cos \left( {{x^2} + 1} \right)}}{h}\]
Hence, after expanding the square, we can write
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {\left( {{x^2} + 1} \right) + \left( {2xh + {h^2}} \right)} \right) - \cos \left( {{x^2} + 1} \right)}}{h}\]
Using the identity \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\] , we can also write it as,
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {{x^2} + 1} \right)\cos \left( {2xh + {h^2}} \right) - \sin \left( {{x^2} + 1} \right)\sin \left( {2xh + {h^2}} \right) - \cos \left( {{x^2} + 1} \right)}}{h}\]
Taking common part aside, that gives,
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \left( {{x^2} + 1} \right)\dfrac{{\cos \left( {2xh + {h^2}} \right) - 1}}{h} - \sin \left( {{x^2} + 1} \right)\dfrac{{\sin \left( {2xh + {h^2}} \right)}}{h}\]
Multiplying and dividing by \[\left( {2x + h} \right)\] , we get
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \left( {{x^2} + 1} \right)\dfrac{{\cosh \left( {2x + h} \right) - 1}}{{h\left( {2x + h} \right)}}\left( {2x + h} \right) - \sin \left( {{x^2} + 1} \right)\dfrac{{\sinh \left( {2x + h} \right)}}{{h\left( {2x + h} \right)}}\left( {2x + h} \right)\] ...(1)
Now since we have,
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\cosh \left( {2x + h} \right) - 1}}{{h\left( {2x + h} \right)}}\left( {2x + h} \right) = \mathop {\lim }\limits_{t \to 0} \dfrac{{\cos t - 1}}{t} = 0\]
And
\[ \Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \left( {2x + h} \right)}}{{h\left( {2x + h} \right)}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sin t}}{t} = 1\]
Hence, putting these values in the equation (1) gives us
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \left( {{x^2} + 1} \right)\dfrac{{\cosh \left( {2x + h} \right) - 1}}{{h\left( {2x + h} \right)}}\left( {2x + h} \right) - \sin \left( {{x^2} + 1} \right)\dfrac{{\sinh \left( {2x + h} \right)}}{{h\left( {2x + h} \right)}}\left( {2x + h} \right)\]
Or,
\[ \Rightarrow f'\left( x \right) = \cos \left( {{x^2} + 1} \right) \cdot 0 \cdot \left( {2x + 0} \right) - \sin \left( {{x^2} + 1} \right) \cdot 1 \cdot \left( {2x + 0} \right)\]
Hence,
\[ \Rightarrow f'\left( x \right) = 0 - \sin \left( {{x^2} + 1} \right) \cdot 2x\]
That gives us finally the derivative as,
\[ \Rightarrow f'\left( x \right) = - \sin \left( {{x^2} + 1} \right) \cdot 2x\]
That is the required derivative.
Therefore, the derivative of \[\cos \left( {{x^2} + 1} \right)\] using the first principle of derivative is \[ - \sin \left( {{x^2} + 1} \right) \cdot 2x\] .

Note: Alternatively, we could have also found the derivative of \[\cos \left( {{x^2} + 1} \right)\] using another well known method that is known as chain rule. It is much quicker and easier compared to the long method of the first principle of derivative.
According to the chain rule, the derivative of a function of the form \[y = f\left( {g\left( x \right)} \right)\] is given by,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
Where \[u = g\left( x \right)\] .