Question

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

Answer 1

Hint: To find the derivative of $-\sin \left( x \right)$ , we have to find the derivative of $\sin x$ and then multiply it with -1. We will first equate $f\left( x \right)=\sin x$ and find the derivative using the formula $f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . We will use $\sin a-\sin b=2\sin \dfrac{1}{2}\left( a-b \right)\cos \dfrac{1}{2}\left( a+b \right)$ and $\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1$ and simplify further. Then, we will use the properties of limits and apply the limits. Finally, we will multiply the derivative with -1.

Complete step by step solution:
We have to find the derivative of $-\sin \left( x \right)$ . Let us first find the derivative of $\sin x$ and then multiply it with -1.
We know that derivative of a function $f\left( x \right)$ is given by
$f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Let us consider $f\left( x \right)=\sin x$ . Then the above formula becomes
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( x+h \right)-\sin x}{h}$
We know that $\sin a-\sin b=2\sin \dfrac{1}{2}\left( a-b \right)\cos \dfrac{1}{2}\left( a+b \right)$ . Let us use this formula in the above equation.
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \dfrac{1}{2}\left( x+h-x \right)\cos \dfrac{1}{2}\left( x+h+x \right)}{h}$
We can simplify the above equation to
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2x+h}{2} \right)}{h}$
We can rewrite the angle of cos in the above expression as
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2x}{2}+\dfrac{h}{2} \right)}{h} \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \left( \dfrac{h}{2} \right)\cos \left( x+\dfrac{h}{2} \right)}{h} \\
\end{align}$
We know that $\displaystyle \lim_{x\to a}f\left( x \right)\cdot g\left( x \right)=\displaystyle \lim_{x\to a}f\left( x \right)\cdot \displaystyle \lim_{x\to a}g\left( x \right)$ . Hence, we can write the above equation as
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2\sin \left( \dfrac{h}{2} \right)}{h}\cdot \displaystyle \lim_{h \to 0}\cos \left( x+\dfrac{h}{2} \right)$
We can rewrite the above equation as
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}\cdot \displaystyle \lim_{h \to 0}\cos \left( x+\dfrac{h}{2} \right)$
We know that $\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1$ . Hence, the above equation becomes
$\begin{align}
  & \Rightarrow f'\left( x \right)=1\cdot \displaystyle \lim_{h \to 0}\cos \left( x+\dfrac{h}{2} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\cos \left( x+\dfrac{h}{2} \right) \\
\end{align}$
Let us apply the limit.
$\begin{align}
  & \Rightarrow f'\left( x \right)=\cos \left( x+0 \right) \\
 & \Rightarrow f'\left( x \right)=\cos x \\
\end{align}$
Now, let us multiply -1 with $f\left( x \right)$ .
$\begin{align}
  & \Rightarrow f\left( x \right)\times -1=\sin x \\
 & \Rightarrow f\left( x \right)=-\sin x \\
\end{align}$
We can write the derivative as
$\begin{align}
  & \Rightarrow f'\left( x \right)\times -1=\cos x \\
 & \Rightarrow f'\left( x \right)=-\cos x \\
\end{align}$

Hence, the derivative of $-\sin \left( x \right)$ is $-\cos x$

Note: Students must know the trigonometric properties and formulas to solve this problem. They must also know the formula of derivatives, the properties of limits and how to apply them. Students must be careful with the formula $\sin a-\sin b=2\sin \dfrac{1}{2}\left( a-b \right)\cos \dfrac{1}{2}\left( a+b \right)$ . The cos part has $a+b$ not $a-b$ . We can also find the derivative of $-\sin \left( x \right)$ in an alternate way.
We know that derivative of $\sin x$ , that is $\dfrac{d}{dx}\sin x=\cos x$ .
Now, we will multiply -1 on both the sides in the above formula.
$\begin{align}
  & -1\times \dfrac{d}{dx}\sin x=\left( \cos x \right)\times -1 \\
 & \Rightarrow \dfrac{d}{dx}\left( -\sin x \right)=-\cos x \\
\end{align}$