MHB 7.6.16 Find the derivative of y with respect to x

[karush]

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

didn't get the arctan ?

 

[MarkFL]

$\tiny{7.6.16}$
$\textsf{Find the derivative of y with respect to x}$
\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 $x$?

 

[karush]

what is

w.r.t

 

[HallsofIvy]

what is

w.r.t

"With respect to".

"What do you get when you implicitly differentiate with respect to x?"

 

[MarkFL]

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

 

[karush]

I was referring to the implicit differentiation of:

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

:D

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

 

[MarkFL]

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

Recall:

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