Question

Differentiate tan2x from first principle
[a] ${{\sec }^{2}}x$
[b] ${{\sec }^{2}}2x$
[c] $1-{{\sec }^{2}}2x$
[d] $2{{\sec }^{2}}2x$

Answer 1

- Hint: Use the fact that $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Take f(x)= cos2x and write $\tan x$ as $\dfrac{\sin x}{\cos x}$ and hence simplify the expression $\tan \left( 2x+2h \right)-\tan 2x$. Use $\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$. Use $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( h \right)}{h}=1$. 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 $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Taking f(x) = tan2x, we get
$ f'\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 $\tan x=\dfrac{\sin x}{\cos x}$.
Hence, we have
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sin \left( 2x+2h \right)}{\cos \left( 2x+2h \right)}-\dfrac{\sin 2x}{\cos 2x}}{h}$
Multiplying the numerator and the denominator by $\cos \left( 2x+2h \right)\cos 2x$, we get
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2x+2h \right)\cos 2x-\sin 2x\cos \left( 2x+2h \right)}{h\cos 2x\cos \left( 2x+2h \right)}$
We know that sinAcosB-cosAsinB = sin(A-B)
Put A = 2x+2h and B = 2x, we get
$\sin \left( 2x+2h \right)\cos 2x-\cos \left( 2x+2h \right)\sin 2x=\sin \left( 2x+2h-2x \right)=\sin 2h$
Hence, we have
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( 2h \right)}{h\cos 2x\cos \left( 2x+2h \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos 2x}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos \left( 2x+2h \right)}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{h}$
We know that $ \underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos 2x}=\dfrac{1}{\cos 2x}=\sec 2x $ (because x is independent of h) and $ \underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\cos \left( 2x+2h \right)}=\dfrac{1}{\cos \left( 2x+0 \right)}=\sec 2x $ (because $\cos 2x\ne 0$ in the domain of tan2x).
Also, we have $ \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{2h}=1 $
Multiplying both sides by 2, we get
$ \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin 2h}{h}=2 $
Hence, we have $f'\left( x \right)=\sec 2x\times \sec 2x\times 2=2{{\sec }^{2}}2x$
Hence the derivative of tan2x is $2{{\sec }^{2}}2x$
Hence option [d] is correct.

Note: [1] Verification using chain rule:
We know that $\dfrac{d}{dx}\left( fog\left( x \right) \right)=\dfrac{d}{d\left( g\left( x \right) \right)}f\left( g\left( x \right) \right)\dfrac{d}{dx}g\left( x \right)$
Taking f(x) = tanx and g(x) = 2x, we have fog(x) = tan2x.
Now, we have $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$
Hence, we have $\dfrac{d}{d\left( g\left( x \right) \right)}\tan \left( g\left( x \right) \right)={{\sec }^{2}}\left( g\left( x \right) \right)\Rightarrow \dfrac{d}{d\left( 2x \right)}\tan \left( 2x \right)={{\sec }^{2}}2x$
We know that $\dfrac{d}{dx}2x=2\dfrac{d}{dx}x=2$
Hence, we have
$\dfrac{d}{dx}\tan \left( 2x \right)={{\sec }^{2}}2x\left( 2 \right)=2{{\sec }^{2}}2x$, which is the same as obtained above.
[2] You can also solve the above question using $\tan \left( 2x+2h \right)=\dfrac{\tan 2x+\tan 2h}{1-\tan 2x\tan 2h}$ and then simplifying and using $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan \left( h \right)}{h}=1$ as shown below:
We have $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\tan 2x+\tan 2h}{1-\tan 2x\tan 2h}-\tan 2x}{h}$
Hence, we have
$\begin{align}
  & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan 2x+\tan 2h-\tan 2x+{{\tan }^{2}}2x\tan 2h}{h\left( 1-\tan 2x\tan 2h \right)} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan 2h}{h}\left( 1+{{\tan }^{2}}2x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{1-\tan 2x\tan 2h} \\
 & =2{{\sec }^{2}}2x \\
\end{align}$
Which is the same as obtained above.