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Question: Differentiate tan2x from first principle [a] \({{\sec }^{2}}x\) [b] \({{\sec }^{2}}2x\) [c] \(...

Differentiate tan2x from first principle
[a] sec⁡2x{{\sec }^{2}}x
[b] sec⁡22x{{\sec }^{2}}2x
[c] 1−sec⁡22x1-{{\sec }^{2}}2x
[d] 2sec⁡22x2{{\sec }^{2}}2x

Explanation

Solution

- Hint: Use the fact that 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}. Take f(x)= cos2x and write tan⁡x\tan x as sin⁡xcos⁡x\dfrac{\sin x}{\cos x} and hence simplify the expression tan⁡(2x+2h)−tan⁡2x\tan \left( 2x+2h \right)-\tan 2x. Use sin⁡Acos⁡B−cos⁡Asin⁡B=sin⁡(A−B)\sin A\cos B-\cos A\sin B=\sin \left( A-B \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 value of the limit and hence find the derivative. Verify your answer using the chain rule of differentiation.

Complete step-by-step solution -

We know that 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}
Taking f(x) = tan2x, we get
f′(x)=lim⁡h→0 f(x+h)−f(x)h=lim⁡h→0 tan⁡(2(x+h))−tan⁡(2x)hf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( 2\left( x+h \right) \right)-\tan \left( 2x \right)}{h}
We know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
Hence, we have
f′(x)=lim⁡h→0 sin⁡(2x+2h)cos⁡(2x+2h)−sin⁡2xcos⁡2xhf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sin \left( 2x+2h \right)}{\cos \left( 2x+2h \right)}-\dfrac{\sin 2x}{\cos 2x}}{h}
Multiplying the numerator and the denominator by cos⁡(2x+2h)cos⁡2x\cos \left( 2x+2h \right)\cos 2x, we get
f′(x)=lim⁡h→0 sin⁡(2x+2h)cos⁡2x−sin⁡2xcos⁡(2x+2h)hcos⁡2xcos⁡(2x+2h)f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2x+2h \right)\cos 2x-\sin 2x\cos \left( 2x+2h \right)}{h\cos 2x\cos \left( 2x+2h \right)}
We know that sinAcosB-cosAsinB = sin(A-B)
Put A = 2x+2h and B = 2x, we get
sin⁡(2x+2h)cos⁡2x−cos⁡(2x+2h)sin⁡2x=sin⁡(2x+2h−2x)=sin⁡2h\sin \left( 2x+2h \right)\cos 2x-\cos \left( 2x+2h \right)\sin 2x=\sin \left( 2x+2h-2x \right)=\sin 2h
Hence, we have
f′(x)=lim⁡h→0 sin⁡(2h)hcos⁡2xcos⁡(2x+2h)=lim⁡h→0 1cos⁡2xlim⁡h→0 1cos⁡(2x+2h)lim⁡h→0 sin⁡2hhf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2h \right)}{h\cos 2x\cos \left( 2x+2h \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos 2x}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos \left( 2x+2h \right)}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{h}
We know that lim⁡h→0 1cos⁡2x=1cos⁡2x=sec⁡2x\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos 2x}=\dfrac{1}{\cos 2x}=\sec 2x (because x is independent of h) and lim⁡h→0 1cos⁡(2x+2h)=1cos⁡(2x+0)=sec⁡2x\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos \left( 2x+2h \right)}=\dfrac{1}{\cos \left( 2x+0 \right)}=\sec 2x (because cos⁡2x≠0\cos 2x\ne 0 in the domain of tan2x).
Also, we have lim⁡h→0 sin⁡2h2h=1\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{2h}=1
Multiplying both sides by 2, we get
lim⁡h→0 sin⁡2hh=2\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{h}=2
Hence, we have f′(x)=sec⁡2x×sec⁡2x×2=2sec⁡22xf'\left( x \right)=\sec 2x\times \sec 2x\times 2=2{{\sec }^{2}}2x
Hence the derivative of tan2x is 2sec⁡22x2{{\sec }^{2}}2x
Hence option [d] is correct.

Note: [1] Verification using chain rule:
We know that ddx(fog(x))=dd(g(x))f(g(x))ddxg(x)\dfrac{d}{dx}\left( fog\left( x \right) \right)=\dfrac{d}{d\left( g\left( x \right) \right)}f\left( g\left( x \right) \right)\dfrac{d}{dx}g\left( x \right)
Taking f(x) = tanx and g(x) = 2x, we have fog(x) = tan2x.
Now, we have ddxtan⁡x=sec⁡2x\dfrac{d}{dx}\tan x={{\sec }^{2}}x
Hence, we have dd(g(x))tan⁡(g(x))=sec⁡2(g(x))⇒dd(2x)tan⁡(2x)=sec⁡22x\dfrac{d}{d\left( g\left( x \right) \right)}\tan \left( g\left( x \right) \right)={{\sec }^{2}}\left( g\left( x \right) \right)\Rightarrow \dfrac{d}{d\left( 2x \right)}\tan \left( 2x \right)={{\sec }^{2}}2x
We know that ddx2x=2ddxx=2\dfrac{d}{dx}2x=2\dfrac{d}{dx}x=2
Hence, we have
ddxtan⁡(2x)=sec⁡22x(2)=2sec⁡22x\dfrac{d}{dx}\tan \left( 2x \right)={{\sec }^{2}}2x\left( 2 \right)=2{{\sec }^{2}}2x, which is the same as obtained above.
[2] You can also solve the above question using tan⁡(2x+2h)=tan⁡2x+tan⁡2h1−tan⁡2xtan⁡2h\tan \left( 2x+2h \right)=\dfrac{\tan 2x+\tan 2h}{1-\tan 2x\tan 2h} and then simplifying and using lim⁡h→0 tan⁡(h)h=1\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( h \right)}{h}=1 as shown below:
We have f′(x)=lim⁡h→0 tan⁡2x+tan⁡2h1−tan⁡2xtan⁡2h−tan⁡2xhf'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\tan 2x+\tan 2h}{1-\tan 2x\tan 2h}-\tan 2x}{h}
Hence, we have
f′(x)=lim⁡h→0 tan⁡2x+tan⁡2h−tan⁡2x+tan⁡22xtan⁡2hh(1−tan⁡2xtan⁡2h) =lim⁡h→0 tan⁡2hh(1+tan⁡22x)lim⁡h→0 11−tan⁡2xtan⁡2h =2sec⁡22x \begin{aligned} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan 2x+\tan 2h-\tan 2x+{{\tan }^{2}}2x\tan 2h}{h\left( 1-\tan 2x\tan 2h \right)} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan 2h}{h}\left( 1+{{\tan }^{2}}2x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{1-\tan 2x\tan 2h} \\\ & =2{{\sec }^{2}}2x \\\ \end{aligned}
Which is the same as obtained above.