# Math Help - fine the derivative

1. ## fine the derivative

$d/dx [(tanx)^{secx}]$

do i use substition or the chain rule?

2. ## use chain rule

and the fact that $\frac{d}{dx} a^x = ln(a)a^x$

3. Originally Posted by zeg
and the fact that $\frac{d}{dx} a^x = ln(a)a^x$
so that is equal to $ln(tanx)tanx^{secx}$, what am i suppose to do now

4. ## my post was incorrect, sorry

that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
$y = tan(x)^{sec(x)}$
$ln(y) = sec(x)\cdot ln(tan(x))$
now take your derivative, d/dx
$(ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}$
$\frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)$

so

$y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}$

that seems messy though. maybe there's some nice simplifications in there somewhere.

5. Originally Posted by zeg
that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
$y = tan(x)^{sec(x)}$
$ln(y) = sec(x)\cdot ln(tan(x))$
now take your derivative, d/dx
$(ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}$
$\frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)$

so

$y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}$

that seems messy though. maybe there's some nice simplifications in there somewhere.

will this work, since $d/dx(p^x) = p^xlnp$, then we have $tanx^{secx}lntanx = tanx^{secx}/tanx$ but i dont know what $tanx^{secx}$ is

6. ## i'm not sure how to simplify that any further

i'm not sure how to simplify that any further. Maybe use the Euler relations?

$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$
$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

if you can simplify $tan(x)^{sec(x)}$ with these relations, then I would do that first, then take the derivative. sorry I can't be more helpful!

7. Originally Posted by zeg
i'm not sure how to simplify that any further. Maybe use the Euler relations?

$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$
$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

if you can simplify $tan(x)^{sec(x)}$ with these relations, then I would do that first, then take the derivative. sorry I can't be more helpful!

thx, you have gave a great help already, don't worry about it

8. Originally Posted by zeg
that is for when a is a constant in this case, it is a function of x, so here's my stab at it:
$y = tan(x)^{sec(x)}$
$ln(y) = sec(x)\cdot ln(tan(x))$
now take your derivative, d/dx
$(ln(y))^{\prime} = (sec(x))^{\prime}\cdot ln(tan(x)) + sec(x)\cdot (ln(tan(x)))^{\prime}$
$\frac{y^{\prime}}{y} = sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)$

so

$y^{\prime} = [sec(x)\cdot tan(x)\cdot ln(tan(x)) + csc(x)]\cdot tan(x)^{sec(x)}$

that seems messy though. maybe there's some nice simplifications in there somewhere.

I think you made a mistake here, (lntan(x))' = csc x sec x, so that is equal to $cscx sec^2x$