Derivative of x^tan(x)

**Bean** · September 12th 2012, 05:31 PM

I thought I knew how to do this but I keep coming up with the wrong answer. Here is what I've done:

$y=x^\tan(x)$

$ln y= tan(x) ln x$

$\frac{1}{y}*\frac{dy}{dx}= \tan(x)*\frac{1}{x} +ln x*sec^2 x$

Then multiply with the original y to get:

$\frac{dy}{dx}= x^{\tan(x)} * (\frac{tan(x)}{x}+ ln x*sec^2 x)$

But the answer I'm seeing has $x^{\tan(x)-1}$ in there instead of the original y value.

Any help would be appreciated.

**topsquark** · September 12th 2012, 05:50 PM

Originally Posted by Bean

I thought I knew how to do this but I keep coming up with the wrong answer. Here is what I've done:

$y=x^\tan(x)$

$ln y= tan(x) ln x$

$\frac{1}{y}*\frac{dy}{dx}= \tan(x)*\frac{1}{x} +ln x*sec^2 x$

Then multiply with the original y to get:

$\frac{dy}{dx}= x^{\tan(x)} * (\frac{tan(x)}{x}+ ln x*sec^2 x)$

But the answer I'm seeing has $x^{\tan(x)-1}$ in there instead of the original y value.

Any help would be appreciated.

It looks to me that if we take a 1/x out of both terms then you would have the correct exponential. But since there are two terms in the parenthesis you'd have to take the 1/x out of both terms.

-Dan

**SworD** · September 12th 2012, 05:55 PM

I would suggest that instead of taking the log, you change the function to the following form:

$y = e^{tan(x) \cdot ln(x)}$ , which is equivalent because $a^{bc} = (a^b)^c$ .

Then, use the chain rule to find the derivative:

$y' = e^{tan(x) \cdot ln(x)} \cdot \left (\frac{\tan(x)}{x} + \sec^2(x)\cdot \ln(x)\right )$

$y' = x^{\tan(x)} \cdot \left (\frac{\tan(x)}{x} + \sec^2(x)\cdot \ln(x)\right )$

Edit: actually, your answer is correct. Perhaps the reason for the discrepancy is that they distributed and changed

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

These are equivalent because $a^{b+c} = a^b \cdot a^c$ .

**Bean** · September 12th 2012, 06:13 PM

Oh right. I see now that the second term was changed to $x sec^{2} x ln x$ in the solution. Thanks, guys.

**Soroban** · September 12th 2012, 07:31 PM

Hello, Bean!

I thought I knew how to do this but I keep coming up with the wrong answer.

Here is what I've done:

$y\:=\:x^\tan(x)$

$\ln y\:=\: \tan(x) \ln x$

$\frac{1}{y}\cdot\frac{dy}{dx}\:=\: \tan x\cdot \frac{1}{x} + \ln x\cdot\sec^2 x$

Then multiply with the original y to get:

$\frac{dy}{dx}\:=\: x^{\tan x}\cdot \left(\tfrac{\tan x}{x}+ \ln x\!\cdot\!\sec^2 x\right)$ . This is correct!

But the answer I'm seeing has: $x^{\tan x-1}$ in there instead of the original y value.

They did something sneaky . . . (legal but sneaky!)

We have: . $\frac{dy}{dx} \;=\;\underbrace{x^{\tan x}\!\cdot\!\frac{\tan x}{x}} \;+\; x^{\tan x}\!\cdot\!\ln x\!\cdot\!\sec^2x$

. . Note that: . $x^{\tan x}\!\cdot\!\frac{\tan x}{x} \;=\;\frac{x^{\tan x}}{x}\!\cdot\!\tan x \;=\; x^{\tan x -1}\!\cdot\!\tan x$

"Subtract exponents", remember?

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