Question

Find the derivatives of the following:

${\displaystyle {\tan }^{-1}=\left(\frac{6x}{1-9x^2}\right)}$

Answer 1

Let y = ${\displaystyle {\tan }^{-1}\left(\frac{6x}{1-9x^2}\right)}$

y = ${\displaystyle {\tan }^{-1}\left(\frac{2\times 3x}{1-{\left(3x\right)}^2}\right)}$  ........(1)

Put 3x = tan θ

(1) ⇒ y = ${\displaystyle {\tan }^{-1}\left(\frac{2\times \tan \theta }{1-{\tan }^2\theta }\right)}$

 y = ${\displaystyle {\tan }^{-1}\left(\tan 2\theta \right)}$

y = 2θ

${\displaystyle \frac{\text{d}y}{\text{d}\theta }=2 \times 1}$ = 2 ........(2)

3x = tan θ

${\displaystyle 3\frac{\text{d}x}{\text{d}\theta }={\sec }^2\theta }$

${\displaystyle 3\frac{\text{d}x}{\text{d}\theta }=1+{\tan }^2\theta }$

${\displaystyle \frac{\text{d}x}{\text{d}\theta }=\frac{1}{3}(1+{\left(3x\right)}^2)}$

${\displaystyle \frac{\text{d}x}{\text{d}\theta }=\frac{1}{3}(1+9x^2)}$  .........(3)

Using equation (2) and (3) we have

${\displaystyle \frac{\frac{\text{d}y}{\text{d}\theta }}{\frac{\text{d}x}{\text{d}\theta }}=\frac{2}{\frac{1}{3}(1+9x^2)}}$

${\displaystyle \frac{\text{d}y}{\text{d}x}=\frac{6}{1+9x^2}}$