Question

Let $ f\left( x \right)=\tan 2x.\tan 3x.\tan 5x $ , then find the value of $ {{f}^{'}}\left( \pi \right) $ .
A. 10
B. -10
C. 0
D. 1

Answer 1

Hint: We first try to explain the concept of the chain rule. We take arbitrary functions and find the differentiation of functions as the form of multiplications. We assume the given functions and apply the chain rule twice to find the differentiated form. Then we put the value of $ x=\pi $ to find the value of $ {{f}^{'}}\left( \pi \right) $ .

Complete step by step answer:
For our given problem we first need to find the differentiation of the given problem $ f\left( x \right)=\tan 2x.\tan 3x.\tan 5x $ . We differentiate both sides of the equations with respect to x.
We apply the theorem of chain rule where $ f\left( x \right)=g\left( x \right).h\left( x \right) $ , the differentiation of the function is $ {{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right).h\left( x \right)+g\left( x \right).{{h}^{'}}\left( x \right) $ . Here $ {{f}^{'}}\left( x \right),{{g}^{'}}\left( x \right),{{h}^{'}}\left( x \right) $ represents the differentiation of functions $ f\left( x \right),g\left( x \right),h\left( x \right) $ with respect to x.
For $ f\left( x \right)=\tan 2x.\tan 3x.\tan 5x $ , we assume $ g\left( x \right)=\tan 2x.\tan 3x,h\left( x \right)=\tan 5x $ .
We know the differential form of $ \tan mx $ is \[\dfrac{d\left( \tan mx \right)}{dx}=m{{\sec }^{2}}mx\].
We first try to find the differentiation of $ g\left( x \right)=\tan 2x.\tan 3x $ . Applying chain rule, we get \[{{g}^{'}}\left( x \right)=\tan 2x.\dfrac{d\left( \tan 3x \right)}{dx}+\dfrac{d\left( \tan 2x \right)}{dx}.\tan 3x=\tan 2x\left( 3{{\sec }^{2}}3x \right)+\left( 2{{\sec }^{2}}2x \right)\tan 3x\]
Now we go back for the differentiation of $ f\left( x \right)=\tan 2x.\tan 3x.\tan 5x $ .
 $ {{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right).h\left( x \right)+g\left( x \right).{{h}^{'}}\left( x \right) $ . Placing all the values we get
 $ {{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right)\left( \tan 5x \right)+g\left( x \right).\dfrac{d\left( \tan 5x \right)}{dx} $ .
We have both the values of \[{{g}^{'}}\left( x \right)\] and the differentiation. We put the values
\[\begin{align}
  & {{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right)\left( \tan 5x \right)+g\left( x \right).\dfrac{d\left( \tan 5x \right)}{dx} \\
 & =\left[ \tan 2x\left( 3{{\sec }^{2}}3x \right)+\left( 2{{\sec }^{2}}2x \right)\tan 3x \right]\left( \tan 5x \right)+\left( \tan 2x.\tan 3x \right)\left( 5{{\sec }^{2}}5x \right) \\
 & =3\tan 2x\tan 5x{{\sec }^{2}}3x+2\tan 3x\tan 5x{{\sec }^{2}}2x+5\tan 2x\tan 3x{{\sec }^{2}}5x \\
\end{align}\]
Now we need to find the value of $ {{f}^{'}}\left( \pi \right) $ . We put the value $ x=\pi $ .
 $ {{f}^{'}}\left( \pi \right)=3\tan 2\pi \tan 5\pi {{\sec }^{2}}3\pi +2\tan 3\pi \tan 5\pi {{\sec }^{2}}2\pi +5\tan 2\pi \tan 3\pi {{\sec }^{2}}5\pi $ .
We have the identity value of $ \tan \left( m\pi \right) $ as 0, $ \forall m\in \mathbb{Z} $ . So, $ \tan 2\pi =\tan 3\pi =\tan 5\pi =0 $ .
Now we have the value of $ {{f}^{'}}\left( \pi \right) $ .
 $ {{f}^{'}}\left( \pi \right)=3\tan 2\pi \tan 5\pi {{\sec }^{2}}3\pi +2\tan 3\pi \tan 5\pi {{\sec }^{2}}2\pi +5\tan 2\pi \tan 3\pi {{\sec }^{2}}5\pi =0 $ .
The value of $ {{f}^{'}}\left( \pi \right) $ is 0. The correct option is C.

Note:
 We can see that in case of three functions being in multiplied form where we have to apply the chain rule, we can apply the form for individual functions and keep the rest of the functions intact. So, the final differentiation formula for $ f\left( x \right)=g\left( x \right).h\left( x \right).k\left( x \right) $ becomes $ {{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right).h\left( x \right).k\left( x \right)+g\left( x \right).k\left( x \right).{{h}^{'}}\left( x \right)+g\left( x \right).h\left( x \right).{{k}^{'}}\left( x \right) $ .