2

I was trying to differentiate the question and I did it in the following 2 ways:

METHOD $1:$

Using the cain rule, we get, $$\frac{\,d}{\,dx}\mathrm{e}^{x\arctan\left(x\right)}=\mathrm{e}^{x\arctan\left(x\right)}\frac{\,d}{\,dx}x\arctan\left(x\right)=\mathrm{e}^{x\arctan\left(x\right)}\left\{\arctan\left(x\right)+\frac{x}{x^2+1}\right\}\tag1$$

METHOD $2:$

But, when we try to do implicit differentiation, we get $$y=\mathrm{e}^{x\arctan\left(x\right)}$$

$$\ln y=x\arctan\left(x\right)$$

$$\frac{\,d}{\,dx}\tan\left(\frac{\ln\left(y\right)}{x}\right)=\frac{\,d}{\,dx}x$$

$$\frac{1}{\cos^2\left(\frac{\ln\left(y\right)}{x}\right)}\frac{1}{xy}y’=1$$

$$y’=\cos^2\left(\tan^{-1}\left(x\right)\right)x\mathrm{e}^{x\arctan\left(x\right)}$$

$$y’=\frac{x}{x^2+1}\mathrm{e}^{x\arctan\left(x\right)}$$

But these give different answers, please help me where have I gone wrong.

Asv
  • 500
  • You did not use the product rule in the fourth line of your second method while differentiating $\ln y/x$. – Karthik Kannan Jun 17 '20 at 13:36
  • In lines 3 to 4 of your second method, you have to use the quotient rule to differentiate $\ln(y) / x$. Another way is just to take the derivative on both sides in line 2 and resubstitute $y$. – paulinho Jun 17 '20 at 13:38
  • @paulinho, I did that, that is how I got $\frac{1}{xy}$ – Asv Jun 17 '20 at 13:39
  • Can you show the steps? I believe there should be an $\ln(y)$ in there somewhere: https://www.wolframalpha.com/input/?i=d%2Fdx+%28ln%28y%28x%29%29%2Fx%29. – paulinho Jun 17 '20 at 15:05

1 Answers1

2

For the implicit differentiation: $$\frac {y'}{y} = \arctan(x) + \frac{x}{x^2+1} \\ y' = e^{x\arctan(x)} \left( \arctan(x) + \frac{x}{x^2+1} \right)$$

Aniruddha Deb
  • 4,345
  • 11
  • 30