Derivative of inverse tangent function

[karush]

Find the derivative of the function $f(y)$
$$f(y)=\tan^{-1}\left({8{y}^{3}+1}\right)$$

 

[Greg]

$$y=\tan^{-1}(x)$$

$$\tan(y)=x$$

$$y'\sec^2(y)=1$$

$$y'=\dfrac{1}{\sec^2(y)}$$

$$y'=\dfrac{1}{1+\tan^2(y)}$$

$$y'=\dfrac{1}{1+x^2}$$

For your problem (chain rule):

$$\dfrac{df(g(y))}{dy}=\dfrac{g'(y)}{1+(g(y))^2}$$

 

[Evgeny.Makarov]

Some explanation to these formulas would be helpful to someone who is "clueless", for example, how you derive $y'\sec^2(y)=1$ from $\tan(y)=x$. Also, what exactly are $f$ and $y$ in the last equation?
\[
\dfrac{df(y)}{dy}=\dfrac{f'(y)}{1+(f(y))^2}
\]
Do you mean the following?
\[
\dfrac{d\tan^{-1}(f(y))}{dy}=\dfrac{f'(y)}{1+(f(y))^2}
\]
But then this redefines $f$ because in post #1 $f(y)=\tan^{-1}(8y^3+1)$ and here $f(y)=8y^3+1$.

karush, you can refer to the list of derivatives of trigonometric functions in Wikipedia. This article explains how to find $(\tan^{-1}(x))'$. In the future, it would help if you explain what exactly you don't understand. In this case, it is not clear whether you know the chain rule or you need the derivative of $\tan^{-1}(x)$ and, if you need the latter, whether you can look it up or need to derive it as a derivative of inverse function.

 

[Greg]

Yes; I've edited my post to reflect your correction. Thanks for the catch!

 

[karush]

$$ \frac{8y'{y}^{3}+{y}^{'}}{(8y^2 +1)^2+1} =
\dfrac{24y^2}{\left(8y^3+1\right)^2+1}
$$

 

[Evgeny.Makarov]

$$ \frac{8y'{y}^{3}+{y}^{'}}{(8y^2 +1)^2+1} =
\dfrac{24y^2}{\left(8y^3+1\right)^2+1}
$$

The right-hand side is a correct answer; the left-hand side is not.

 

[karush]

thus ??
$$ \dfrac{\class{steps-node}{\cssId{steps-node-4}{8\cdot\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}y}\left[y^3\right]}}+\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}y}\left[1\right]}}}}}{\left(8y^3+1\right)^2+1}
=
\dfrac{24y^2}{\left(8y^3+1\right)^2+1}
$$

 

[Evgeny.Makarov]

Yes, I don't know how else you could get the right-hand side in post #5.

 

[karush]

I thoght $y'$ Was the same as $\d{}{y}$

 

[topsquark]

I thoght $y'$ Was the same as $\d{}{y}$

No. y is still a function of x. Thus y' = dy/dx

-Dan

 

[Evgeny.Makarov]

In post #1 $y$ looks like an independent variable.