1. ## Trigo derivative urgent help please...

what is the derivative of (sinx)^tanx??

2. logarithmic differentiation ...

$\displaystyle y = (\sin{x})^{\tan{x}}$

$\displaystyle \ln(y) = \ln\left[(\sin{x})^{\tan{x}}\right]$

$\displaystyle \ln(y) = \tan{x} \cdot \ln(\sin{x})$

take the derivative w/r to x ...

$\displaystyle \frac{y'}{y} = \tan{x} \cdot \frac{\cos{x}}{\sin{x}} + \ln(\sin{x}) \cdot \sec^2{x}$

$\displaystyle \frac{y'}{y} = 1 + \sec^2{x} \cdot \ln(\sin{x})$

$\displaystyle y' = (\sin{x})^{\tan{x}}\left[1 + \sec^2{x} \cdot \ln(\sin{x})\right]$

3. Originally Posted by juneau
what is the derivative of (sinx)^tanx??
$\displaystyle sin(x)^{tan(x)} = y$

$\displaystyle log(y) = tan(x)log(sin(x))$

$\displaystyle \frac{1}{y} \frac{dy}{dx} = sec^2(x)log(sin(x)) + tan(x)\times cos(x)\times \frac{1}{sin(x)}$ ------Using Chain Rule

$\displaystyle \frac{dy}{dx} = y\times (sec^2(x)log(sin(x)) + 1)$

If you want put the value of y in RHS that's your answer
EDIT: Skeeter won

4. well see, the original question was: y = (x)^(sinx)^(tanx)

I already used log differentiation to get (sinx)^tanx down. can I use log differentiation again??

5. Originally Posted by juneau
well see, the original question was: y = (x)^(sinx)^(tanx)

I already used log differentiation to get (sinx)^tanx down. can I use log differentiation again??
why did you not mention that to begin with?

$\displaystyle y = x^{\sin{x}^{\tan{x}}}$

$\displaystyle \ln{y} = \ln\left(x^{\sin{x}^{\tan{x}}}\right)$

$\displaystyle \ln{y} = \sin{x}^{\tan{x}} \cdot \ln{x}$

use the product rule ...

$\displaystyle \frac{y'}{y} = \sin{x}^{\tan{x}} \cdot \frac{1}{x} + \ln{x} \cdot (derivative \, from \, my \, previous \, post)$

finish up.

6. Yes, you can, using the chain rule.

By the way, here's an interesting point: if $\displaystyle y= f(x)^{g(x)}$, where f and g are integrable functions, there are two common mistakes in finding the derivative:

1) Ignore the fact that g is a variable and use the power law:
$\displaystyle y'= g(x) f(x)^{g(x)-1}$

2) Ignore the fact that f is a variable and treat it as an exponential:
$\displaystyle y'= ln(f(x)) f(x)^{g(x)}$

As I said, those are both wrong. The correct derivative is the sum of those!

$\displaystyle y'= g(x) f(x)^{g(x)-1}+ ln(f(x)) f(x)^{g(x)}$