Question

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

Answer 1

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:
$\dfrac{d}{{dx}}\sin x = \cos x$,
$\dfrac{d}{{dx}}\cos x = - \sin x$,
$\dfrac{d}{{dx}}\tan x = {\sec ^2}x$,
$\dfrac{d}{{dx}}\cot x = - {\csc ^2}x$,
$\dfrac{d}{{dx}}\sec x = \sec x\tan x$,
$\dfrac{d}{{dx}} = - {\csc ^2}x\cot x$.