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Question: What is the derivative of \(-\sin \left( x \right)\) ?...

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

Explanation

Solution

To find the derivative of −sin⁡(x)-\sin \left( x \right) , we have to find the derivative of sin⁡x\sin x and then multiply it with -1. We will first equate f(x)=sin⁡xf\left( x \right)=\sin x and find the derivative using the formula f′(x)=lim⁡h→0f(x+h)−f(x)hf'\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=2sin⁡12(a−b)cos⁡12(a+b)\sin a-\sin b=2\sin \dfrac{1}{2}\left( a-b \right)\cos \dfrac{1}{2}\left( a+b \right) and lim⁡x→0sin⁡xx=1\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⁡(x)-\sin \left( x \right) . Let us first find the derivative of sin⁡x\sin x and then multiply it with -1.
We know that derivative of a function f(x)f\left( x \right) is given by
f′(x)=lim⁡h→0f(x+h)−f(x)hf'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
Let us consider f(x)=sin⁡xf\left( x \right)=\sin x . Then the above formula becomes
⇒f′(x)=lim⁡h→0sin⁡(x+h)−sin⁡xh\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=2sin⁡12(a−b)cos⁡12(a+b)\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.
⇒f′(x)=lim⁡h→02sin⁡12(x+h−x)cos⁡12(x+h+x)h\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
⇒f′(x)=lim⁡h→02sin⁡(h2)cos⁡(2x+h2)h\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
⇒f′(x)=lim⁡h→02sin⁡(h2)cos⁡(2x2+h2)h ⇒f′(x)=lim⁡h→02sin⁡(h2)cos⁡(x+h2)h \begin{aligned} & \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{aligned}
We know that lim⁡x→af(x)⋅g(x)=lim⁡x→af(x)⋅lim⁡x→ag(x)\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
⇒f′(x)=lim⁡h→02sin⁡(h2)h⋅lim⁡h→0cos⁡(x+h2)\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
⇒f′(x)=lim⁡h→0sin⁡(h2)h2⋅lim⁡h→0cos⁡(x+h2)\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 lim⁡x→0sin⁡xx=1\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1 . Hence, the above equation becomes
⇒f′(x)=1⋅lim⁡h→0cos⁡(x+h2) ⇒f′(x)=lim⁡h→0cos⁡(x+h2) \begin{aligned} & \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{aligned}
Let us apply the limit.
⇒f′(x)=cos⁡(x+0) ⇒f′(x)=cos⁡x \begin{aligned} & \Rightarrow f'\left( x \right)=\cos \left( x+0 \right) \\\ & \Rightarrow f'\left( x \right)=\cos x \\\ \end{aligned}
Now, let us multiply -1 with f(x)f\left( x \right) .
⇒f(x)×−1=sin⁡x ⇒f(x)=−sin⁡x \begin{aligned} & \Rightarrow f\left( x \right)\times -1=\sin x \\\ & \Rightarrow f\left( x \right)=-\sin x \\\ \end{aligned}
We can write the derivative as
⇒f′(x)×−1=cos⁡x ⇒f′(x)=−cos⁡x \begin{aligned} & \Rightarrow f'\left( x \right)\times -1=\cos x \\\ & \Rightarrow f'\left( x \right)=-\cos x \\\ \end{aligned}

Hence, the derivative of −sin⁡(x)-\sin \left( x \right) is −cos⁡x-\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=2sin⁡12(a−b)cos⁡12(a+b)\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+ba+b not a−ba-b . We can also find the derivative of −sin⁡(x)-\sin \left( x \right) in an alternate way.
We know that derivative of sin⁡x\sin x , that is ddxsin⁡x=cos⁡x\dfrac{d}{dx}\sin x=\cos x .
Now, we will multiply -1 on both the sides in the above formula.
−1×ddxsin⁡x=(cos⁡x)×−1 ⇒ddx(−sin⁡x)=−cos⁡x \begin{aligned} & -1\times \dfrac{d}{dx}\sin x=\left( \cos x \right)\times -1 \\\ & \Rightarrow \dfrac{d}{dx}\left( -\sin x \right)=-\cos x \\\ \end{aligned}