Question

If $y=\tan x\cdot \tan 2x\cdot \tan 3x$ , then $\dfrac{dy}{dx}$ has the value equal to
(a) $6{{\sec }^{2}}3x\cdot \tan x\cdot \tan 2x+2{{\sec }^{2}}x\cdot \tan 2x\cdot \tan 3x+4{{\sec }^{2}}2x\cdot \tan 3x\cdot \tan x$
(b) $2y\left( \csc 2x+2\csc 4x+3\csc 6x \right)$
(c) $3{{\sec }^{2}}3x-2{{\sec }^{2}}2x-{{\sec }^{2}}x$
(d) ${{\sec }^{2}}x+2{{\sec }^{2}}2x+3{{\sec }^{2}}3x$

Answer 1

Hint: First, we will expand the last term i.e. $\tan 3x$ as $\tan 3x=\tan \left( 2x+x \right)$ . Then we will apply the formula $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . By solving this equation, we will find an equation, something like this $y=\tan 3x-\tan 2x-\tan x$ . After this, we have to differentiate the equation and we will find the answer. Also, we will use the formula $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\cdot 1$ .

Complete step-by-step solution -
Here, we are given the equation $y=\tan x\cdot \tan 2x\cdot \tan 3x$ . So, first we will expand the term $\tan 3x$ as $\tan 3x=\tan \left( 2x+x \right)$ .
This is similar to the formula $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . Here $a=2x,b=x$ . So, using this formula and putting values, we get
$\Rightarrow \tan 3x=\tan \left( 2x+x \right)$
$\Rightarrow \tan 3x=\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$
On further solving, we get equation as
$\Rightarrow \tan 3x\left( 1-\tan 2x\tan x \right)=\tan 2x+\tan x$
Now, we will multiply the brackets and we get
$\Rightarrow \tan 3x-\tan x\tan 2x\tan 3x=\tan 2x+\tan x$
We are given in the question that $\tan x\tan 2x\tan 3x=y$ so, substituting the value we get
$\Rightarrow \tan 3x-y=\tan 2x+\tan x$
Now, rearranging the terms we get
$\Rightarrow y=\tan 3x-\tan 2x-\tan x$ ……………………………….(1)
Now, we will differentiate the equation (1) and we will get as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan 3x \right)-\dfrac{d}{dx}\left( \tan 2x \right)-\dfrac{d}{dx}\left( \tan x \right)$
Now, we know the formula that $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\cdot 1$ because coefficient of x is 1 so, multiplying it with 1. So, using this formula we get
$\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}3x\cdot 3-{{\sec }^{2}}2x\cdot 2-{{\sec }^{2}}x\cdot 1$
$\Rightarrow \dfrac{dy}{dx}=3{{\sec }^{2}}3x-2{{\sec }^{2}}2x-{{\sec }^{2}}x$
Thus, option (c) is the correct answer.

Note: Another approach of solving this problem is using product rule i.e. given as $\dfrac{d}{dx}\left( uv \right)=u\cdot \dfrac{d}{dx}v+v\cdot \dfrac{d}{dx}u$ . By using this, we will be able to solve the first 2 terms but then multiplying it with the third term we will again make it more complex and time consuming. But at the end, we will get the same answer. So, it is better to go with the solution given which is simpler rather than this product rule formula.