Question

Find the derivative of sin2x.

Answer 1

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

Complete step-by-step solution -

Let $f\left( x \right)=\sin 2x$
Hence we have
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
f(x+h) = sin(2(x+h))
Hence we have
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2\left( x+h \right) \right)-\sin 2x}{h}$
We know that $\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)$
Replace x by 2x+2h and y by 2x, we get
$\sin \left( 2x+2h \right)-\sin 2x=2\cos \left( \dfrac{2x+2h+2x}{2} \right)\sin \left( \dfrac{2x+2h-2x}{2} \right)=2\cos \left( 2x+h \right)\sin \left( h \right)$
Hence we have
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left( 2x+h \right)\sin \left( h \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,2\cos \left( 2x+h \right)\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}$
We know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$.
Using the above result, we get
$f'\left( x \right)=2\cos \left( 2x+0 \right)1=2\cos 2x$
Hence we have
$\dfrac{d}{dx}\sin 2x=2\cos 2x$

Note: [1] Alternatively, we have
Let f(x) = sinx and g(x) = 2x.
Then we have h(x) = fog(x) = sin2x.
We know that $\dfrac{d}{dx}fog\left( x \right)=\dfrac{d}{d\left( g\left( x \right) \right)}fog\left( x \right)\dfrac{d}{dx}g\left( x \right)$. This is known as the chain rule of differentiation.
Now we know that $\dfrac{d}{dx}\sin x=\cos x$
Hence we have
$\dfrac{d}{d\left( 2x \right)}\sin \left( 2x \right)=\cos 2x$
Also, we have $\dfrac{d}{dx}2x=2$
Hence we have from chain rule of differentiation
$\dfrac{d}{dx}\sin 2x=\dfrac{d}{d\left( 2x \right)}\sin \left( 2x \right)\dfrac{d}{dx}2x=\cos 2x\left( 2 \right)=2\cos 2x$, 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
$\dfrac{d}{dx}\left( \sin 2x \right)=\dfrac{d}{dx}\left( 2\sin x\cos x \right)=2\dfrac{d}{dx}\left( \sin x\cos x \right)$
Using product rule of differentiation, we have
$\dfrac{d}{dx}\sin 2x=2\sin x\dfrac{d}{dx}\cos x+2\cos x\dfrac{d}{dx}\sin x$
Now, we know that the derivative of sinx is cosx and the derivative of cosx is -sinx.
Hence we have
$\dfrac{d}{dx}\sin 2x=2\sin x\left( -\sin x \right)+2\cos x\cos x=2\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$
Hence, we have
$\dfrac{d}{dx}\sin 2x=2\left( \cos 2x \right)=2\cos 2x$, which is the same as obtained above