How do you differentiate \[\cos \left( { - x} \right)\]?

Hint: In solving the question, first assume the given expression to a variable \[y\], i.e.,\[y = \cos \left( { - x} \right)\], now differentiate this equation by using derivatives of trigonometry functions and by using chain rule, we will get the required result.

Complete step-by-step solution:
Differentiation can be defined as a derivative of independent variable value and can be used to calculate features in an independent variable per unit modification.
The derivative of any function \[y = f\left( x \right)\] with respect to variable \[x\] is measure of the rate at which the value of the function changes with respect to the change in the value of variable \[x\]. The first derivative of any function also signifies the slope of the function when the graph of \[x\] considers only real values of the function.
Now assume the given expression as variable\[y\], we will get,\[y = f\left( x \right)\]is plotted against
\[y = \cos \left( { - x} \right)\],
Now applying differentiation with implicit differentiation on the Left hand side and chain rule on the right hand side we get,
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\cos \left( { - x} \right)\],
Now using chain rule we get,
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\cos ( - x) \cdot \dfrac{d}{{dx}}\left( x \right)\],
Now differentiating we get,
\[\dfrac{{dy}}{{dx}} = \-sin ( - x) \cdot -1\],
Now simplifying we get,
\[\dfrac{{dy}}{{dx}} = \sin ( - x)\],
We know that \[\sin \left( { - x} \right) = - \sin x\].
So the derivative of the given expression \[\cos \left( { - x} \right)\] is \[ - \sin x\].

\[\therefore \]The differentiation value of \[\cos \left( { - x} \right)\] is \[ - \sin x\].

Note: Differentiation is the method of evaluating a function’s derivative at any time. The definition of trigonometry is the interaction of angles and triangle faces. We have 6 major ratios here, they are sine, cosine, tangent, cotangent, secant, and cosecant. Some of the derivatives of trigonometric functions are given below:
\[\dfrac{d}{{dx}}\sin x = \cos x\],
\[\dfrac{d}{{dx}}\cos x = - \sin x\],
\[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\],
\[\dfrac{d}{{dx}}\cot x = - {\csc ^2}x\],
\[\dfrac{d}{{dx}}\sec x = \sec x\tan x\],
\[\dfrac{d}{{dx}} = - {\csc ^2}x\cot x\].