Question

What is the derivative of $\cos \left( -x \right)$?

Answer 1

In this problem we need to calculate the derivative of the given trigonometric function. We can observe the given trigonometric ratio which is $\cos \left( -x \right)$. So, we will first use the negative angle trigonometric formula which is $\cos \left( -\theta \right)=\cos \theta$ to simplify the given value $\cos \left( -x \right)$. Now we will differentiate the obtained equation with respect to $x$. After that we will calculate the required result by applying differentiation formulas.

Complete step-by-step answer:
Given value is $\cos \left( -x \right)$.
We can observe a negative angle in the above given trigonometric ratio. So, we are going to use the negative angle trigonometric formula which is $\cos \left( -\theta \right)=\cos \theta$ for the given value, then the given value is modified as
$\cos \left( -x \right)=\cos x$
Differentiating the above equation with respect to the variable $x$, then we will get
$\dfrac{d}{dx}\left( \cos \left( -x \right) \right)=\dfrac{d}{dx}\left( \cos x \right)$
We have the differentiation formula that is $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Using this formula in the above equation, then we will have
$\dfrac{d}{dx}\left( \cos \left( -x \right) \right)=-\sin x$

Hence the derivative of the given value $\cos \left( -x \right)$ is $-\sin x$.

Note: We can also use another method to solve the above equation. We can use the substitution method to find the derivative of the given value. Use the substitution $y=-x$ in the given value, then we will get
$\cos \left( -x \right)=\cos y$
Differentiate the equation $y=-x$ with respect to $x$, then we will get
$\begin{aligned} & \dfrac{dy}{dx}=-\dfrac{dx}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=-1 \\\ \end{aligned}$
Now differentiating the equation $\cos \left( -x \right)=\cos y$ with respect to variable $x$, then we will have
$\dfrac{d}{dx}\left( \cos \left( -x \right) \right)=\dfrac{d}{dx}\left( \cos y \right)$
Using the formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ in the above equation, then we will get
$\dfrac{d}{dx}\left( \cos \left( -x \right) \right)=-\sin y\times \dfrac{dy}{dx}$
Substituting the values $y=-x$, $\dfrac{dy}{dx}=-1$ in the above equation, then we will have
$\begin{aligned} & \dfrac{d}{dx}\left( \cos \left( -x \right) \right)=-\sin \left( -x \right)\times \left( -1 \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( \cos \left( -x \right) \right)=\sin \left( -x \right) \\\ \end{aligned}$
Using the formula $\sin \left( -x \right)=-\sin x$ in the above equation, then we will get
$\dfrac{d}{dx}\left( \cos \left( -x \right) \right)=-\sin x$
From both the methods we got the same result but the method which is discussed earlier is simple and doesn't have so many substitutions.