Find the derivative at the point $(1, \ln(\pi/4))$ of $x \arctan x = e^y$.
By Sophia Bowman
Find the derivative at the point $(1, \ln(\pi/4)$ of $x \arctan x = e^y$.
So I think this may be an implicit differentiation problem. I'm not exactly sure where to start. I tried differentiating it with respect to $y$, but that won't work. $\ln(x \arctan x) = y$?
Thanks in advance.
$\endgroup$ 64 Answers
$\begingroup$Instead of using implicit differentiation, a simpler approach is to take the natural log of both sides then differentiate.
$$x\cdot\arctan x=e^y \rightarrow \ln(x\cdot\arctan x) = y$$
Now just differentiate, using the chain and product rule
\begin{align} \frac{dy}{dx} & = \frac{1}{x\cdot\arctan x} \cdot \left(\arctan x \space + \space \frac{x}{1 + x^2}\right) \\ \end{align}
Plugging in the point $\left(1, \ln (\pi/4)\right)$ yields the following
\begin{align} f'(1) & =\frac{1}{1\cdot\arctan 1} \cdot \left(\arctan 1 \space + \space \frac{1}{1 + 1^2}\right) \\ & = \frac{1}{\frac{\pi}{4}} \cdot \left( \frac{\pi}{4} + \frac{1}{2}\right) \\ & = 1 + \frac{2}{\pi} \end{align}
$\endgroup$ $\begingroup$Hint: $\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}$ and $\frac{d}{dx}\ln x=\frac{1}{x}$
$\endgroup$ $\begingroup$This is not an answer but it is too long for a comment.
As Noble Mushtak answered, the problem with $x \tan^{-1} (x) = e^y$ is quite simple. Let us make it more complex and consider $$x \tan^{-1} (x) = f(y)$$ where $f(y)$ is any function of $y$.
In this case, as you suspected, you would have needed the differentiation of the implicit function $$F=x \tan^{-1} (x) - f(y)=0$$ The partial derivatives being $$F'_x=\frac{x}{x^2+1}+\tan ^{-1}(x)$$ $$F'_y=-f'(y)$$ using the implicit function theorem leads to $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{\frac{x}{x^2+1}+\tan ^{-1}(x)} {F'_y}$$
$\endgroup$ $\begingroup$You need to differentiate both sides with respect to $x$ to find the derivative, which almost certainly refers to $\frac{dy}{dx}$. When we solve for $\frac{dy}{dx}$, we will get both $x$ and $y$ in the derivative. Once we have $\frac{dy}{dx}$, we need to substitute $x=1$ and $y=\ln\frac{\pi}{4}$ to find the derivative at that point.
Now, here is our equation: $$x\cdot\arctan x=e^y$$ Take the derivative of both sides: $$\frac{d}{dx}(x\cdot\arctan x)=\frac{d}{dx}(e^y)$$ Let's do the right side first since it's shorter. Here, we use chain rule. First, we take the derivative of $e^y$ with respect to $y$ to get $e^y$ and then we multiply by the derivative of $y$ with respect to $x$, which is $\frac{dy}{dx}$. This gives us: $$\frac{d}{dx}(x\cdot\arctan x)=e^y\cdot\frac{dy}{dx}$$
Now, on the left side, we use product rule: $$\frac{d}{dx}(x)\cdot\arctan x+\frac{d}{dx}(\arctan x)\cdot x=e^y\cdot\frac{dy}{dx}$$ $$\arctan x+\frac{x}{x^2+1}=e^y\cdot\frac{dy}{dx}$$ Now, we want to solve for $\frac{dy}{dx}$, so we divide both sides by $e^y$ and switch the sides to get: $$\frac{dy}{dx}=\frac{\arctan x+\frac{x}{x^2+1}}{e^y}$$
Finally, we need to find the derivative at $(1, \ln \frac\pi 4)$. This means $x=1$ and $y=\ln \frac\pi 4$, so: $$\frac{dy}{dx}=\frac{\arctan 1+\frac{1}{1^2+1}}{e^{\ln \frac\pi 4}}$$ $$\frac{dy}{dx}=\frac{\frac\pi 4+\frac 1 2}{\frac\pi 4}$$ $$\frac{dy}{dx}=1+\frac{2}{\pi}$$
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