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Question: Find the derivative of sin2x....

Find the derivative of sin2x.

Explanation

Solution

Hint: Use the definition of derivative of a function f′(x)=lim⁡h→0 f(x+h)−f(x)hf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}. Use the fact that sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right). Use lim⁡h→0 sin⁡(h)h=1\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}=1. Hence find the derivative of sin2x. Alternative use chain rule of differentiation, i.e. ddx(f(g(x)))=dd(g(x))f((g(x)))ddxg(x)\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=\dfrac{d}{d\left( g\left( x \right) \right)}f\left( \left( g\left( x \right) \right) \right)\dfrac{d}{dx}g\left( x \right)
Use the fact that ddxsin⁡x=cos⁡x\dfrac{d}{dx}\sin x=\cos x and ddxx=1\dfrac{d}{dx}x=1.

Complete step-by-step solution -

Let f(x)=sin⁡2xf\left( x \right)=\sin 2x
Hence we have
f′(x)=lim⁡h→0 f(x+h)−f(x)hf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}
f(x+h) = sin(2(x+h))
Hence we have
f′(x)=lim⁡h→0 sin⁡(2(x+h))−sin⁡2xhf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2\left( x+h \right) \right)-\sin 2x}{h}
We know that sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)
Replace x by 2x+2h and y by 2x, we get
sin⁡(2x+2h)−sin⁡2x=2cos⁡(2x+2h+2x2)sin⁡(2x+2h−2x2)=2cos⁡(2x+h)sin⁡(h)\sin \left( 2x+2h \right)-\sin 2x=2\cos \left( \dfrac{2x+2h+2x}{2} \right)\sin \left( \dfrac{2x+2h-2x}{2} \right)=2\cos \left( 2x+h \right)\sin \left( h \right)
Hence we have
f′(x)=lim⁡h→0 2cos⁡(2x+h)sin⁡(h)h=lim⁡h→0 2cos⁡(2x+h)×lim⁡h→0 sin⁡(h)hf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left( 2x+h \right)\sin \left( h \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,2\cos \left( 2x+h \right)\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}
We know that lim⁡x→0 sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1.
Using the above result, we get
f′(x)=2cos⁡(2x+0)1=2cos⁡2xf'\left( x \right)=2\cos \left( 2x+0 \right)1=2\cos 2x
Hence we have
ddxsin⁡2x=2cos⁡2x\dfrac{d}{dx}\sin 2x=2\cos 2x

Note: [1] Alternatively, we have
Let f(x) = sinx and g(x) = 2x.
Then we have h(x) = fog(x) = sin2x.
We know that ddxfog(x)=dd(g(x))fog(x)ddxg(x)\dfrac{d}{dx}fog\left( x \right)=\dfrac{d}{d\left( g\left( x \right) \right)}fog\left( x \right)\dfrac{d}{dx}g\left( x \right). This is known as the chain rule of differentiation.
Now we know that ddxsin⁡x=cos⁡x\dfrac{d}{dx}\sin x=\cos x
Hence we have
dd(2x)sin⁡(2x)=cos⁡2x\dfrac{d}{d\left( 2x \right)}\sin \left( 2x \right)=\cos 2x
Also, we have ddx2x=2\dfrac{d}{dx}2x=2
Hence we have from chain rule of differentiation
ddxsin⁡2x=dd(2x)sin⁡(2x)ddx2x=cos⁡2x(2)=2cos⁡2x\dfrac{d}{dx}\sin 2x=\dfrac{d}{d\left( 2x \right)}\sin \left( 2x \right)\dfrac{d}{dx}2x=\cos 2x\left( 2 \right)=2\cos 2x, which is same as obtained above
[2] Alternative Solution 2:
We know that sin2x =2sinxcosx and (uv)’=u’v+v’u {This is known as product rule of differentiation}
Hence we have
ddx(sin⁡2x)=ddx(2sin⁡xcos⁡x)=2ddx(sin⁡xcos⁡x)\dfrac{d}{dx}\left( \sin 2x \right)=\dfrac{d}{dx}\left( 2\sin x\cos x \right)=2\dfrac{d}{dx}\left( \sin x\cos x \right)
Using product rule of differentiation, we have
ddxsin⁡2x=2sin⁡xddxcos⁡x+2cos⁡xddxsin⁡x\dfrac{d}{dx}\sin 2x=2\sin x\dfrac{d}{dx}\cos x+2\cos x\dfrac{d}{dx}\sin x
Now, we know that the derivative of sinx is cosx and the derivative of cosx is -sinx.
Hence we have
ddxsin⁡2x=2sin⁡x(−sin⁡x)+2cos⁡xcos⁡x=2(cos⁡2x−sin⁡2x)\dfrac{d}{dx}\sin 2x=2\sin x\left( -\sin x \right)+2\cos x\cos x=2\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)
We know that cos⁡2x=cos⁡2x−sin⁡2x\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x
Hence, we have
ddxsin⁡2x=2(cos⁡2x)=2cos⁡2x\dfrac{d}{dx}\sin 2x=2\left( \cos 2x \right)=2\cos 2x, which is the same as obtained above