# Thread: differentiation of trig functions in mod

1. ## differentiation of trig functions in mod

The Question and My Attempt in the pic

i got correct answer but i think my method has some flaw ...

2. It's fine.

3. Can you help me to do this one ...

4. You will need to consider how you can rewrite $\displaystyle |x|$ and $\displaystyle |\sin{x}|$ in the neighbourhood of $\displaystyle x = -\frac{\pi}{4}$.

5. I gues both |x| and |sin x| will be negative so |x| = -x and |sin x| = -sin x

the i can take log on f(x)

that way i'm getting something weird .... like (pi/4)^(1/√2) * [4/√2 pi]

6. I get $\left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}}\ left(-\dfrac{\ln{\dfrac{\pi}{4}}}{\sqrt{2}} - \dfrac{2\sqrt{2}}{\pi}\right)$.

$f(x) = \left(-x\right)^{-\sin{x}} = e^{\ln{\left(-x\right)^{-\sin{x}}}} = e^{-\sin{x}\ln{\left(-x\right)}}$

Let $u = -\sin{x}\ln{\left(-x\right)}$. Then, $du = -\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}$.

Substitute $u = -\sin{x}\ln{\left(-x\right)}$ in f(x).

$f(x) = e^u$.

Then, f'(x) is easy to compute.

$f'(x) = e^u du = e^{-\sin{x}\ln{\left(-x\right)}} \left(-\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}\right) = \left(-x\right)^{-\sin{x}} \left(-\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}\right)$

Substitute $x = -\dfrac{\pi}{4}$ in f'(x).

$f'(-\dfrac{\pi}{4}) = \left(--\dfrac{\pi}{4}\right)^{-\sin{\left(-\dfrac{\pi}{4}\right)}} \left(-\cos{\left(-\dfrac{\pi}{4}\right)}\ln{\left(--\dfrac{\pi}{4}\right)} - \dfrac{\sin{\left(-\dfrac{\pi}{4}\right)}}{-\dfrac{\pi}{4}}\right)$

Get rid of the double negatives.

$f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{-\sin{\left(-\dfrac{\pi}{4}\right)}} \left(-\cos{\left(-\dfrac{\pi}{4}\right)}\ln{\left(\dfrac{\pi}{4}\rig ht)} + \dfrac{\sin{\left(-\dfrac{\pi}{4}\right)}}{\dfrac{\pi}{4}}\right)$

Next, get rid of the sines and cosines: $sin{\left(-\dfrac{\pi}{4}\right)} = -\dfrac{1}{\sqrt{2}}$ and $cos{\left(-\dfrac{\pi}{4}\right)} = \dfrac{1}{\sqrt{2}}$.

$f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{--\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right) } + \dfrac{-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}\right)$

Get rid of double negatives.

$f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right) } + \dfrac{-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}\right)$

Get rid of the fraction within a fraction within a fraction within a fraction.

$f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right) } - \dfrac{4}{\sqrt{2}\pi}\right) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right) } - \dfrac{4\sqrt{2}}{2\pi}\right) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right) } - \dfrac{2\sqrt{2}}{\pi}\right)$

Edit: Prove It's method seems less convoluted. :P

7. Well let's see...

I agree that you would rewrite this as $\displaystyle y = (-x)^{-\sin{x}}$.

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

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

$\displaystyle \frac{d}{dx}(\ln{y}) = \frac{d}{dx}\left[-\sin{x}\ln{(-x)}\right]$

$\displaystyle \frac{1}{y}\,\frac{dy}{dx} = -\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}$

$\displaystyle \frac{1}{(-x)^{-\sin{x}}}\,\frac{dy}{dx} = -\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}$

$\displaystyle \frac{dy}{dx} = (-x)^{-\sin{x}}\left[-\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}\right]$.

When this is evaluated at $\displaystyle x = -\frac{\pi}{4}$

$\displaystyle \frac{dy}{dx} = \left(\frac{\pi}{4}\right)^{\frac{\sqrt{2}}{2}}\le ft[-\frac{2\sqrt{2}}{\pi} - \frac{\sqrt{2}}{2}\ln{\left(\frac{\pi}{4}\right)}\ right]$.

8. Hi guys ...

really thanks for the replies ... i really appreciate it ......

i am not getting how it will be 4/pi and how will the log term be positive ???

really thanks for help

9. $\displaystyle -\frac{\sqrt{2}}{2}\ln{\left(\frac{\pi}{4}\right)} = \frac{\sqrt{2}}{2}\ln{\left[\left(\frac{\pi}{4}\right)^{-1}\right]} = \frac{\sqrt{2}}{2}\ln{\left(\frac{4}{\pi}\right)}$

10. OH YES!!!

That was dumb of me

sorry for trouble

Thanks "prove it" for help