7.6.16 Find the derivative of y with respect to x

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Discussion Overview

The discussion revolves around finding the derivative of the function \( y = \ln(\tan^{-1}(4x^3)) \) with respect to \( x \). Participants explore the process of differentiation, including implicit differentiation techniques and the application of the chain rule.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant expresses uncertainty about differentiating the arctangent function.
  • Another participant defines \( u(x) = 4x^3 \) and \( v(x) = \arctan(u(x)) \), suggesting that \( y(x) = \ln(v(x)) \) and proposes using the chain rule for differentiation.
  • There is a request for clarification on the meaning of "with respect to" in the context of differentiation.
  • A participant suggests that the derivative can be expressed as \( \frac{1}{\arctan(4x^3)} \cdot \frac{d}{dx} \arctan(4x^3) \), referencing implicit differentiation.
  • Another participant reiterates the focus on implicit differentiation of the equation \( \tan(v(x)) = u(x) \) and provides a derivative expression involving \( \sec^2(v(x)) \) and \( \frac{d v}{dx} \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivative or the best approach to differentiate the function, with multiple viewpoints and methods being discussed.

Contextual Notes

Some participants reference implicit differentiation and the chain rule, but the discussion includes various interpretations and steps that remain unresolved.

karush
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$\tiny{7.6.16}$
$\textsf{Find the derivative of y with respect to x}$
\begin{align*}\displaystyle
y&=\ln{(\tan^{-1}(4x^3))} \\
y'&=
\end{align*}

didn't get the arctan ?
☕
 
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karush said:
$\tiny{7.6.16}$
$\textsf{Find the derivative of y with respect to x}$
\begin{align*}\displaystyle
y&=\ln{(\tan^{-1}(4x^3))} \\
y'&=
\end{align*}

didn't get the arctan ?
☕

Let's define the following:

$$u(x)=4x^3$$

$$v(x)=\arctan(u(x))$$

And so we have:

$$y(x)=\ln(v(x))$$

Hence:

$$\d{y}{x}=\frac{1}{v(x)}\d{v}{x}$$

Now, we may write:

$$\tan(v(x))=u(x)$$

What do you get when you implicitly differentiate w.r.t $x$?
 
what is

w.r.t
 
karush said:
what is

w.r.t
"With respect to".

"What do you get when you implicitly differentiate with respect to x?"
 
karush said:
this?

$\frac{1}{\arctan\left(4x^3\right)}\cdot \frac{d}{dx}\arctan\left(4x^3\right)$

I was referring to the implicit differentiation of:

$$\tan(v(x))=u(x)$$

:D
 
MarkFL said:
I was referring to the implicit differentiation of:

$$\tan(v(x))=u(x)$$

:D

$$\tan(v(x))=u(x)$$$$\d{v}{x}\tan(v(x))=\d{u}{x}u(x)=12x^2$$
 
karush said:
$$\tan(v(x))=u(x)$$$$\d{v}{x}\tan(v(x))=\d{u}{x}u(x)=12x^2$$

Recall:

$$\frac{d}{dx}\left(\tan(v(x))\right)=\sec^2(v(x))\d{v}{x}$$
 

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